Polycarp wrote on the board a string s containing only lowercase Latin letters (‘a’-‘z’). This string is known for you and given in the input.

After that, he erased some letters from the string s, and he rewrote the remaining letters in any order. As a result, he got some new string t. You have to find it with some additional information.

Suppose that the string t has length m and the characters are numbered from left to right from 1 to m. You are given a sequence of m integers: b1,b2,…,bm, where bi is the sum of the distances |i?j| from the index i to all such indices j that tj>ti (consider that ‘a’<‘b’<…<‘z’). In other words, to calculate bi, Polycarp finds all such indices j that the index j contains a letter that is later in the alphabet than ti and sums all the values |i?j|.

For example, if t = “abzb”, then:

since t1=‘a’, all other indices contain letters which are later in the alphabet, that is: b1=|1?2|+|1?3|+|1?4|=1+2+3=6;
since t2=‘b’, only the index j=3 contains the letter, which is later in the alphabet, that is: b2=|2?3|=1;
since t3=‘z’, then there are no indexes j such that tj>ti, thus b3=0;
since t4=‘b’, only the index j=3 contains the letter, which is later in the alphabet, that is: b4=|4?3|=1.
Thus, if t = “abzb”, then b=[6,1,0,1].

Given the string s and the array b, find any possible string t for which the following two requirements are fulfilled simultaneously:

t is obtained from s by erasing some letters (possibly zero) and then writing the rest in any order;
the array, constructed from the string t according to the rules above, equals to the array b specified in the input data.
Input
The first line contains an integer q (1≤q≤100) — the number of test cases in the test. Then q test cases follow.

Each test case consists of three lines:

the first line contains string s, which has a length from 1 to 50 and consists of lowercase English letters;
the second line contains positive integer m (1≤m≤|s|), where |s| is the length of the string s, and m is the length of the array b;
the third line contains the integers b1,b2,…,bm (0≤bi≤1225).
It is guaranteed that in each test case an answer exists.

Output
Output q lines: the k-th of them should contain the answer (string t) to the k-th test case. It is guaranteed that an answer to each test case exists. If there are several answers, output any.

Example
Input
4
abac
3
2 1 0
abc
1
0
abba
3
1 0 1
ecoosdcefr
10
38 13 24 14 11 5 3 24 17 0
Output
aac
b
aba
codeforces
Note
In the first test case, such strings t are suitable: "aac’, “aab”.

In the second test case, such trings t are suitable: “a”, “b”, “c”.

In the third test case, only the string t equals to “aba” is suitable, but the character ‘b’ can be from the second or third position.

思路：對于一個字符串，肯定有最大的字母和次大的字母等等，對于一開始的b數組來說，如果為0的話，那么這個位置一定是最大字母，然后再減去這個位置對于其他位置的貢獻，下一個為0的位置一定就是次大的字母，一直這樣弄下去直到所有的位置都弄完。
思路很好想，但是實現起來有些困難。
①對于很多個0，怎么計算所有的0的貢獻。
②對于一個字母來說，這個字母用完之后，大于等于這個字母的字母個數都要清零。
這是我遇到的兩個坑點。
代碼如下：

#include<bits/stdc++.h> #define ll long long using namespace std;const int maxx=1e2+10; int vis[maxx],b[maxx]; char s1[maxx]; string s; int n;inline bool cmp(char a,char b) {return a>b; } int main() {int t;scanf("%d",&t);while(t--){cin>>s;scanf("%d",&n);for(int i=0;i<26;i++) vis[i]=0;for(int i=1;i<=n;i++) scanf("%d",&b[i]);for(int i=0;i<s.length();i++) vis[s[i]-'a']++;sort(s.begin(),s.end());int cnt=0;while(cnt<n){int num=0;for(int i=1;i<=n;i++) if(b[i]==0) num++;int pos;for(int i=25;i>=0;i--){if(vis[i]>=num){pos=i;break;}}int rr=num;for(int i=pos;i<26;i++) vis[i]=0;num=0;for(int i=1;i<=n;i++){if(b[i]==0){num+=i;s1[i]=(char)('a'+pos);}}int zz=0,z1=0;for(int i=1;i<=n;i++){if(b[i]==0) zz+=i,z1++,b[i]=-1,cnt++;else if(b[i]==-1) continue;else{b[i]-=((i*z1-zz)+(num-zz)-i*(rr-z1));}}//for(int i=1;i<=n;i++) cout<<b[i]<<" ";cout<<endl;}for(int i=1;i<=n;i++) cout<<s1[i];cout<<endl;}return 0; }

努力加油a啊，(^o)/~

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