題意：

大意就是給一個序列，可能有重復(fù)數(shù)字，有x次機(jī)會為這個序列填上一個數(shù)字，問最終從里面獲得的1~v連續(xù)子序列的v最大是多少。

題目:

Dreamoon is a big fan of the Codeforces contests.

One day, he claimed that he will collect all the places from 1 to 54 after two more rated contests. It’s amazing!

Based on this, you come up with the following problem:

There is a person who participated in n Codeforces rounds. His place in the first round is a1, his place in the second round is a2, …, his place in the n-th round is an.

You are given a positive non-zero integer x.

Please, find the largest v such that this person can collect all the places from 1 to v after x more rated contests.

In other words, you need to find the largest v, such that it is possible, that after x more rated contests, for each 1≤i≤v, there will exist a contest where this person took the i-th place.

For example, if n=6, x=2 and a=[3,1,1,5,7,10] then answer is v=5, because if on the next two contest he will take places 2 and 4, then he will collect all places from 1 to 5, so it is possible to get v=5.

Input

The first line contains an integer t (1≤t≤5) denoting the number of test cases in the input.

Each test case contains two lines. The first line contains two integers n,x (1≤n,x≤100). The second line contains n positive non-zero integers a1,a2,…,an (1≤ai≤100).

Output

For each test case print one line containing the largest v, such that it is possible that after x other contests, for each 1≤i≤v, there will exist a contest where this person took the i-th place.

Example

Input

5
6 2
3 1 1 5 7 10
1 100
100
11 1
1 1 1 1 1 1 1 1 1 1 1
1 1
1
4 57
80 60 40 20

Output

5
101
2
2
60

分析：

其實(shí)讀懂這道題的題意，基本都能寫出來，A題一般都簡單，這里說下我對這道題的分析：
1.我首先想到map，記錄出現(xiàn)的值。
2.用一個while循環(huán)，直到找到第m次的“空位”
3.結(jié)果用了減二，因?yàn)閣hile循環(huán)，先判斷再運(yùn)行再判斷這樣就會多加2
4.起始即為1，題目要求從1開始。

AC代碼：

#include<stdio.h> #include<string.h> #include<map> #include<algorithm> using namespace std; const int M=1e3+10; int t,n,m,ans,k; int a[M]; int main() {scanf("%d",&t);while(t--){map<int,int>mp;scanf("%d%d",&n,&m);for(int i=0;i<n;i++){scanf("%d",&a[i]);mp[a[i]]++;}k=0;ans=1;while(k<=m){if(!mp[ans])k++;ans++;}printf("%d\n",ans-2);} }

In addtion

最近想提高下自己博客的閱讀量，就會開始盡量把以后所有寫過的CodeForces，發(fā)到博客上，因?yàn)轭}會新一點(diǎn)，也希望能幫到你們，我會開一個新分類。就醬！

總結(jié)

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