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.ArrayList;
import java.util.Arrays;
import java.util.*;
public class edu91
{
	public static void main (String[] args)
	 {
		Scanner sc=new Scanner(System.in); 
		int t,i;
		t=sc.nextInt();
		while(t>0)
		{
			t--;
			int n=sc.nextInt();
			int a[]=new int[n];
			for(i=0;i<n;i++)
				a[i]=sc.nextInt();
			int l=-1,j=-1,k=-1;
			String ans="YES";
			for(i=1;i<n;i++)
			{
				if(a[i]>a[i-1])
				{
					l=i;
					while(i<n&&a[i]>a[i-1])
						i++;
					if(i==n)
						ans="NO";
					else
					{
						k=i+1;
						j=i;
					}
					break;
				}
			}
			if(ans.equals("NO")||l==-1||k==-1||j==-1)
				System.out.println("NO");
			else
			{
				System.out.println(ans);
				System.out.println(l+" "+j+" "+k);
			}
		}
	} 
}	