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 ed91_2_1 {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();
		while(t-- > 0) {
			int n =sc.nextInt();
			int a[] = new int[n];
			
			for(int i=0;i<n;i++) {
				a[i] = sc.nextInt();
			}
			
			int x=-1,y=-1,z=-1;
			for(int i=1;i<n-1;i++) {
				if(a[i] > a[i-1] && a[i] > a[i+1]) {
					x=i;
					y=i+1;
					z = i+2;
					break;
				}
			}
			
			if(x==-1 ||y==-1 || z==-1) {
				System.out.println("No");
			}
			else {
				System.out.println("Yes");
				System.out.println(x + " " + y + " " + z);
			}
		}
		sc.close();
	}

}
