A. Do Not Be Distracted!

題意：

一件事情一但開始，只能做完才能做別的事，當(dāng)出現(xiàn)一件事不連續(xù)出現(xiàn)時(shí)，教師會(huì)懷疑

題目：

Polycarp has 26 tasks. Each task is designated by a capital letter of the Latin alphabet.

The teacher asked Polycarp to solve tasks in the following way: if Polycarp began to solve some task, then he must solve it to the end, without being distracted by another task. After switching to another task, Polycarp cannot return to the previous task.

Polycarp can only solve one task during the day. Every day he wrote down what task he solved. Now the teacher wants to know if Polycarp followed his advice.

For example, if Polycarp solved tasks in the following order: “DDBBCCCBBEZ”, then the teacher will see that on the third day Polycarp began to solve the task ‘B’, then on the fifth day he got distracted and began to solve the task ‘C’, on the eighth day Polycarp returned to the task ‘B’. Other examples of when the teacher is suspicious: “BAB”, “AABBCCDDEEBZZ” and “AAAAZAAAAA”.

If Polycarp solved the tasks as follows: “FFGZZZY”, then the teacher cannot have any suspicions. Please note that Polycarp is not obligated to solve all tasks. Other examples of when the teacher doesn’t have any suspicious: “BA”, “AFFFCC” and “YYYYY”.

Help Polycarp find out if his teacher might be suspicious.

Input

The first line contains an integer t (1≤t≤1000). Then t test cases follow.

The first line of each test case contains one integer n (1≤n≤50) — the number of days during which Polycarp solved tasks.

The second line contains a string of length n, consisting of uppercase Latin letters, which is the order in which Polycarp solved the tasks.

Output

For each test case output:

“YES”, if the teacher cannot be suspicious;
“NO”, otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).

Example

input

5
3
ABA
11
DDBBCCCBBEZ
7
FFGZZZY
1
Z
2
AB

output

NO
NO
YES
YES
YES

AC代碼：

#include<stdio.h> #include<string.h> #include<algorithm> #include<string> #include<iostream> using namespace std; typedef long long ll; int t,n,m; int main(){string s;cin>>t;while(t--){cin>>n>>s;int flag=0;int book[30];memset(book,0,sizeof(book));for(int i=0;i<n;i++){while(s[i]==s[i+1])i++;//cout<<<<endl;if(book[s[i]-'A'+1]==1){flag=1;break;}book[s[i]-'A'+1]=1;}flag==1?cout<<"NO"<<endl:cout<<"YES"<<endl;} }

B. Ordinary Numbers

題意：

給你一個(gè)數(shù)n,問（1~n）中有多少個(gè)，每個(gè)數(shù)位都相等的數(shù)？

題目：

Let’s call a positive integer n ordinary if in the decimal notation all its digits are the same. For example, 1, 2 and 99 are ordinary numbers, but 719 and 2021 are not ordinary numbers.

For a given number n, find the number of ordinary numbers among the numbers from 1 to n.

Input

The first line contains one integer t (1≤t≤104). Then t test cases follow.

Each test case is characterized by one integer n (1≤n≤109).

Output

For each test case output the number of ordinary numbers among numbers from 1 to n.

Example

input

6
1
2
3
4
5
100

output

1
2
3
4
5
18

AC代碼：

#include<iostream> using namespace std; int main() {int t;cin>>t;while(t--){long long int n,i,j,count=0;cin>>n;for(i=1; i<=n; i=i*10+1){for(j=1; j<=9; j++)if(i*j<=n){count++;//cout<<i<<"***"<<j<<endl;}}cout<<count<<endl;} }

C. Not Adjacent Matrix

題意：

給你一個(gè)n*n的方格，向里面放置（1~$n?n$）的數(shù)，問如何放，使得任意格子的上下左右格子的差值不為一。

題目：

We will consider the numbers a and b as adjacent if they differ by exactly one, that is, |a?b|=1.

We will consider cells of a square matrix n×n as adjacent if they have a common side, that is, for cell (r,c) cells (r,c?1), (r,c+1), (r?1,c) and (r+1,c) are adjacent to it.

For a given number n, construct a square matrix n×n such that:

Each integer from 1 to n2 occurs in this matrix exactly once;
If (r1,c1) and (r2,c2) are adjacent cells, then the numbers written in them must not be adjacent.

Input

The first line contains one integer t (1≤t≤100). Then t test cases follow.

Each test case is characterized by one integer n (1≤n≤100).

Output

For each test case, output:

-1, if the required matrix does not exist;
the required matrix, otherwise (any such matrix if many of them exist).
The matrix should be outputted as n lines, where each line contains n integers.

Example

input

3
1
2
3

output

1
-1
2 9 7
4 6 3
1 8 5

AC代碼：

#include <bits/stdc++.h> using namespace std;int main() {int t;cin>>t;while(t--){int n;cin>>n;if(n == 2)cout<<"-1\n";else{for(int i=1; i<=n*n; i+=2)//交叉放置即可cout<<i<<' ';for(int i=2; i<=n*n; i+=2)cout<<i<<' ';cout<<endl;}}return 0; }

D. Same Differences

題意：

給你一個(gè)n個(gè)整數(shù)的數(shù)組，問數(shù)組中滿足i<j and $a_j?a_i=j?i，$.有多少個(gè)？

題目：

You are given an array a of n integers. Count the number of pairs of indices (i,j) such that i<j and $a_j?a_i=j?i，$.

Input

The first line contains one integer t (1≤t≤104). Then t test cases follow.

The first line of each test case contains one integer n (1≤n≤2?105).

The second line of each test case contains n integers a1,a2,…,an (1≤ai≤n) — array a.

It is guaranteed that the sum of n over all test cases does not exceed 2?105.

Output

For each test case output the number of pairs of indices (i,j) such that i<j and aj?ai=j?i.

Example

input

4
6
3 5 1 4 6 6
3
1 2 3
4
1 3 3 4
6
1 6 3 4 5 6

output

1
3
3
10

分析：

讓我們重寫一下原始的相等性：
$a_j?a_i=j?i，$
$a_j?j=a_i?i$
讓我們用$b_i=a_i?i$替換每個(gè)$a_i$。 那么答案是對(duì)（i，j）的對(duì)數(shù)，使得i <j并且$b_i=b_j$。 要計(jì)算此值，可以使用地圖或排序。

AC代碼：

#include <bits/stdc++.h> using namespace std; int main() {int t;cin>>t;while(t--){int n;map<int,int> mp;cin>>n;long long ans=0;for(int i=1; i<=n; i++){int x;cin>>x;ans+=mp[x-i];mp[x-i]++;}cout<<ans<<endl;}return 0; }

H - Arranging The Sheep

題意：

給你一個(gè)字符串，表示一排中🐏羊的站位，對(duì)羊進(jìn)行移動(dòng)，使得羊之間沒有空隙，問最小的移動(dòng)數(shù)？

題目：

You are playing the game “Arranging The Sheep”. The goal of this game is to make the sheep line up. The level in the game is described by a string of length n, consisting of the characters ‘.’ (empty space) and ‘*’ (sheep). In one move, you can move any sheep one square to the left or one square to the right, if the corresponding square exists and is empty. The game ends as soon as the sheep are lined up, that is, there should be no empty cells between any sheep.

For example, if n=6 and the level is described by the string “**.*…”, then the following game scenario is possible:

the sheep at the 4 position moves to the right, the state of the level: “….";
the sheep at the 2 position moves to the right, the state of the level: "...";
the sheep at the 1 position moves to the right, the state of the level: "...";
the sheep at the 3 position moves to the right, the state of the level: "...";
the sheep at the 2 position moves to the right, the state of the level: "…*.”;
the sheep are lined up and the game ends.
For a given level, determine the minimum number of moves you need to make to complete the level.

Input

The first line contains one integer t (1≤t≤104). Then t test cases follow.

The first line of each test case contains one integer n (1≤n≤106).

The second line of each test case contains a string of length n, consisting of the characters ‘.’ (empty space) and ‘*’ (sheep) — the description of the level.

It is guaranteed that the sum of n over all test cases does not exceed 106.

Output

For each test case output the minimum number of moves you need to make to complete the level.

Example

Input

5 6 **.*.. 5 ***** 3 .*. 3 ... 10 *.*...*.**

Output

1
0
0
0
9

分析：

因?yàn)樽罱K是讓所有的羊在一起，那么處在最中間的羊，動(dòng)一步，都不是最少的移動(dòng)，所以對(duì)其他羊來說，就都往中間羊靠攏即可。

AC代碼：

#include<bits/stdc++.h> using namespace std; const int M=1e6+10; int a[1000010]; int main() {int t;cin>>t;while(t--){int n;cin>>n;string s;cin>>s;int sum=0;for(int i=0; i<n; i++)if(s[i]=='*')a[sum++]=i;long long ans=0;int m=sum/2;for(int i=0; i<sum; i++)ans+=abs(a[i]-a[m])-abs(i-m);//第i只羊離中間羊的距離；cout<<ans<<endl;}return 0; }

A - Nastia and Nearly Good Numbers

題意：

給你兩個(gè)數(shù)A和B，問是否存在兩個(gè)能夠整除A，不能整除ab的數(shù)x,y，及一個(gè)x+y能夠整除ab的數(shù)z。

題目：

Nastia has 2 positive integers A and B. She defines that:

The integer is good if it is divisible by A?B;
Otherwise, the integer is nearly good, if it is divisible by A.
For example, if A=6 and B=4, the integers 24 and 72 are good, the integers 6, 660 and 12 are nearly good, the integers 16, 7 are neither good nor nearly good.

Find 3 different positive integers x, y, and z such that exactly one of them is good and the other 2 are nearly good, and x+y=z.

Input

The first line contains a single integer t (1≤t≤10000) — the number of test cases.

The first line of each test case contains two integers A and B (1≤A≤106, 1≤B≤106) — numbers that Nastia has.

Output

For each test case print:

“YES” and 3 different positive integers x, y, and z (1≤x,y,z≤1018) such that exactly one of them is good and the other 2 are nearly good, and x+y=z.
“NO” if no answer exists.
You can print each character of “YES” or “NO” in any case.
If there are multiple answers, print any.

Example

Input

3
5 3
13 2
7 11

Output

YES
10 50 60
YES
169 39 208
YES
28 154 182

Note

In the first test case: 60 — good number; 10 and 50 — nearly good numbers.

In the second test case: 208 — good number; 169 and 39 — nearly good numbers.

In the third test case: 154 — good number; 28 and 182 — nearly good numbers.

AC代碼：

#include<stdio.h> #include<string.h> #include<algorithm> #include<iostream> using namespace std; typedef long long ll; ll n,m; int main(){int t;cin>>t;while(t--){cin>>n>>m;if(m==1)cout<<"NO"<<endl;else{cout<<"YES"<<endl;printf("%lld %lld %lld\n",n,m*n,(m+1)*n);//相鄰兩個(gè)數(shù)互質(zhì)}}return 0; }

B - Nastia and a Good Array

題意：

給你一個(gè)包含n個(gè)整數(shù)的數(shù)組，現(xiàn)在經(jīng)過操作：
1.每次選取兩個(gè)整數(shù)，$a_i,a_j$,以及在（1~$2?10^9$）中找到兩個(gè)數(shù)x,y；
2.滿足$min(a_i,a_j)=min(x,y)$
3.最后讓$a_i=x,a_j=y$
使得最終的序列，任意相鄰的兩個(gè)數(shù)都互質(zhì)。

題目·：

Nastia has received an array of n positive integers as a gift.

She calls such an array a good that for all i (2≤i≤n) takes place gcd(ai?1,ai)=1, where gcd(u,v) denotes the greatest common divisor (GCD) of integers u and v.

You can perform the operation: select two different indices i,j (1≤i,j≤n, i≠j) and two integers x,y (1≤x,y≤2?109) so that min(ai,aj)=min(x,y). Then change ai to x and aj to y.

The girl asks you to make the array good using at most n operations.

It can be proven that this is always possible.

Input

The first line contains a single integer t (1≤t≤10000) — the number of test cases.

The first line of each test case contains a single integer n (1≤n≤105) — the length of the array.

The second line of each test case contains n integers a1,a2,…,an (1≤ai≤109) — the array which Nastia has received as a gift.

It’s guaranteed that the sum of n in one test doesn’t exceed 2?105.

Output

For each of t test cases print a single integer k (0≤k≤n) — the number of operations. You don’t need to minimize this number.

In each of the next k lines print 4 integers i, j, x, y (1≤i≠j≤n, 1≤x,y≤2?109) so that min(ai,aj)=min(x,y) — in this manner you replace ai with x and aj with y.

If there are multiple answers, print any.

Example

Input

2
5
9 6 3 11 15
3
7 5 13

Output

2
1 5 11 9
2 5 7 6
0

Note

Consider the first test case.

Initially a=[9,6,3,11,15].

In the first operation replace a1 with 11 and a5 with 9. It’s valid, because min(a1,a5)=min(11,9)=9.

After this a=[11,6,3,11,9].

In the second operation replace a2 with 7 and a5 with 6. It’s valid, because min(a2,a5)=min(7,6)=6.

After this a=[11,7,3,11,6] — a good array.

In the second test case, the initial array is already good.

分析：

由于不要求最優(yōu)，我們可以找每個(gè)間隔都放置一個(gè)質(zhì)數(shù)$1e^9+7$;

AC代碼：

#include<stdio.h> #include<iostream> using namespace std; const int mod=1e9+7; const int M=1e5+10; int a[M]; int main(){int t,n;cin>>t;while(t--){cin>>n;for(int i=1;i<=n;i++){cin>>a[i];}cout<<n/2<<endl;for(int i=1;i<n;i+=2){printf("%d %d %d %d\n",i,i+1,min(a[i],a[i+1]),mod);}}return 0; }

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