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
T = int(input())
for case in range(T):
    n = int(input())
    value = input().split()
    value = [int(val) for val in value]
    v_p_l, v_p_r = [i for i in range(n)], [i for i in range(n)]
    v_p_l[0] = [value[0], 0]
    v_p_r[-1] = [value[-1], n-1]
    decide = False
    ans =[]
    
    for i in range(1, n):
        if value[i] < v_p_l[i-1][0]:
            v_p_l[i] = [value[i], i]
        else:
            v_p_l[i] = v_p_l[i-1]
    for j in range(-2, -n-1, -1):
        if value[j] < v_p_r[j+1][0]:
            v_p_r[j] = [value[j], n+j]
        else:
            v_p_r[j] = v_p_r[j+1]

    for i in range(1, n-1):
        if v_p_l[i-1][0] < value[i] and value[i] > v_p_r[i+1][0]:
            decide = True
            ans = [v+1 for v in [v_p_l[i-1][1], i, v_p_r[i+1][1]]]
            break
            
    if decide is False:
        print("NO")
    else:
        print("YES")
        print("%d %d %d" %(ans[0], ans[1], ans[2]))
