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.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		int m = scan.nextInt();
		while(m-->0) {
			boolean b = false;
			boolean b1 = false;
			int n = scan.nextInt();
			int[]arr = new int[n+1];
			for(int i=1;i<=n;++i) {
				arr[i] = scan.nextInt();
			}
			for(int i=1;i<=n;++i) {
				int j=i;
				for( j=i;j>0;j--) {
					if(arr[i]>arr[j]) {
						b1 = true;
						break;
					}
				}
				if(b1) {
					for(int k=i;k<=n;++k) {
						if(arr[i]>arr[k]) {
							System.out.println("Yes");
							System.out.println(j+" "+i+" "+k);
							b = true;
							break;
						}
					}
				}
				if(b) {
					break;
				}
			
			}
			if(!b) {
				System.out.println("No");
				
			}
				
		}
	}

}
