You are given a permutation p1,p2,…,pnp1,p2,…,pn . Recall that sequence of nn integers is called a permutation if it contains all integers from 11 to nn exactly once.

Find three indices ii , jj and kk such that:

• 1≤i<j<k≤n1≤i<j<k≤n ;
• pi<pjpi<pj and pj>p
  kpj>pk .

Or say that there are no such indices.

Input

The first line contains a single integer TT (1≤T≤2001≤T≤200 )— the number of test cases.

Next 2T2T lines contain test cases— two lines per test case. The first line of each test case contains the single integer nn (3≤n≤10003≤n≤1000 )— the length of the permutation p

The second line contains nn integers p1,p2,…,pnp1,p2,…,pn (1≤pi≤n1≤pi≤n ; pi≠pjpi≠pj if i≠ji≠j )— the permutation pp .

Output

For each test case:

• if there are such indices ii , jj and kk , print YES (case insensitive) and the indices themselves;
• if there are no such indices, print NO (case insensitive).

If there are multiple valid triples of indices, print any of them.

Example

Input

Copy

3

4

2 1 4 3

6

4 6 1 2 5 3

5

5 3 1 2 4

Output

Copy

YES

2 3 4

YES

3 5 6

NO

就…硬暴力。（不過顯然可以優化到O(n)QAQ

#include <bits/stdc++.h>
using namespace std;
int n, a[1005];
int main()
{
    int t;
    cin >> t;
    while(t--){
        cin >> n;
        int x = 0, y = 0, z = 0;
        for(int i = 1; i <= n; i++) cin >> a[i];
        for(int i = 2; i <= n - 1; i++){
            for(int j = 1; j < i; j++){
                if(a[j] < a[i]){
                    x = j;
                    break;
                }
            }
            for(int j = i + 1; j <= n; j++){
                if(a[j] < a[i]){
                    y = j;
                    break;
                }
            }
            if(x && y){
                z = i;
                break;
            }else{
                x = y = z = 0;
            }
        }
        if(z){
            cout << "YES" << endl;
            cout << x << ' ' << z << ' ' << y << endl;
        }else{
            cout << "NO" <<endl;
        }

    }
    //system("pause");
    return 0;
}