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
/*
    https://codeforces.com/problemset/problem/1380/A
 */

import java.util.Scanner;

public class ThreeIndices {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();
        for (int i = 0; i < t; i++) {
            int n = scanner.nextInt();
            int[] arr = new int[n];
            for (int j = 0; j < n; j++) {
                arr[j] = scanner.nextInt();
            }

            int left = 0;
            int right = n - 1;
            int max = n;
            while (left <= right && (max == arr[left] || max == arr[right])) {
                if (max == arr[left]) {
                    left++;
                } else {
                    right--;
                }
                max--;
            }

            if (max == 0) {
                System.out.println("NO");
            } else {
                int maxIndex = -1;
                for (int j = left; j <= right; j++) {
                    if (arr[j] == max) {
                        maxIndex = j;
                        break;
                    }
                }
                System.out.println("YES");
                System.out.format("%d %d %d\n", left + 1, maxIndex + 1, right + 1);
            }
        }
    }
}
