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.Scanner;
public class Main {

    public static void main(String[] args) {
	    Scanner s = new Scanner(System.in);
	    int t = s.nextInt();
	    while(t-->0){
	        int n = s.nextInt();
	        int[] a = new int[n];
	        for(int i=0;i<n;i++){
	            a[i] = s.nextInt();
            }
            ArrayList<Integer> li = new ArrayList<>();
	        int count =0;
	        int diff = 0;
	        boolean v = true;
	        for(int i=0;i<n-1;i++){
	            diff = a[i+1]-a[i];
	            li.add(i);
	            if(diff<0 && count !=0){
	                li.add(i+1);
					System.out.println("YES");
                    System.out.println(li.get(li.size()-3)+1+" " +(li.get(li.size()-2)+1)+" "+(li.get(li.size()-1)+1));
                    v = false;
	                break;
                }
	            else if(diff>=0){
	                count++;
                }
            }
	        if(v){
				System.out.println("NO");
			}
        }
    }
}