Educational Codeforces Round 91 (Rated for Div. 2) A. Three Indices（暴力）

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>pkpj>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 pp .

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。（不过显然可以优化到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;
}