Dreamoon and Ranking Collection CodeForces - 1330A （贪心）

题意：

大意就是给一个序列，可能有重复数字，有x次机会为这个序列填上一个数字，问最终从里面获得的1~v连续子序列的v最大是多少。

题目:

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：

其实读懂这道题的题意，基本都能写出来，A题一般都简单，这里说下我对这道题的分析：
1.我首先想到map，记录出现的值。
2.用一个while循环，直到找到第m次的“空位”
3.结果用了减二，因为while循环，先判断再运行再判断这样就会多加2
4.起始即为1，题目要求从1开始。

AC代码：

#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);}
}