The 16th Zhejiang University Programming Contest-

Handshakes Time Limit: 2 Seconds       Memory Limit: 65536 KB

Last week, n students participated in the annual programming contest of Marjar University. Students are labeled from 1 to n. They came to the competition area one by one, one after another in the increasing order of their label. Each of them went in, and before sitting down at his desk, was greeted by his/her friends who were present in the room by shaking hands.

For each student, you are given the number of students who he/she shook hands with when he/she came in the area. For each student, you need to find the maximum number of friends he/she could possibly have. For the sake of simplicity, you just need to print the maximum value of the n numbers described in the previous line.

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

The first line contains an integer n (1 ≤ n ≤ 100000) -- the number of students. The next line contains n integers a₁, a₂, ..., a_n (0 ≤ a_i < i), where a_i is the number of students who the i-th student shook hands with when he/she came in the area.

Output

For each test case, output an integer denoting the answer.

Sample Input

2
3
0 1 1
5
0 0 1 1 1

Sample Output

2
3

  浙大的校赛题，，不能说坑吧只怪自己理解错了，原来a[i]表示与第i个人进来和a[i]个人握手，然后求朋友最多的数量，可我们做的时候大部分把那句话理解为i与第a[i]个人握手，毕竟a[i]的范围太符合了。。这样理解题意后完全是道水题~~

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
using namespace std;
const int N=100000+10;
int a[N];
int main()
{
    int t,n,i,maxx;
    scanf("%d",&t);
    while(t--)
    {
        maxx=0;
        scanf("%d",&n);
        for(i=1; i<=n; i++)
        {
            scanf("%d",&a[i]);
            if(a[i])
                maxx++;
            if(a[i]>maxx)
                maxx=a[i];
        }
        printf("%d\n",maxx);
    }
    return 0;
}