You are given a permutation p_1, p_2, ..., p_n. Recall that sequence of n integers is called a permutation if it contains all integers from 1 to n exactly once.

Find three indices i, j and k such that: 

  * 1 ≤ i < j < k ≤ n; 
  * p_i < p_j and p_j > p_k. 

Or say that there are no such indices.

Input

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

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

The second line contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n; p_i ≠ p_j if i ≠ j) — the permutation p.

Output

For each test case: 

  * if there are such indices i, j and k, 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


3
4
2 1 4 3
6
4 6 1 2 5 3
5
5 3 1 2 4


Output


YES
2 3 4
YES
3 5 6
NO
import java.util.*;

public class CD1380A {

	public static void main(String[] args) {
		
		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();
	outer:	for(int tt = 0 ; tt < t ;tt++) {
			int n = sc.nextInt();
			int a[] = new int[n];
			for(int i = 0; i < n ;i++) {
				a[i] = sc.nextInt();
			}
			int x = 0;
			int y = -1;
			for(int i = 1 ; i < n ; i++) {
				if(a[i] < a[x] && y== -1) x = i;
				else if(a[i] > a[x] && y == -1) y = i;
				else if(a[i] < a[y]) {
					System.out.println("YES");
					System.out.println((x+1)+" "+(y+1)+" " + (i+1));
					continue outer;
				}
				else y = i;
					
			}System.out.println("NO");
		}
		
		
		sc.close();
		
	}

}
