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
for _ in range(int(input())):
    n = int(input())
    p = list(map(int, input().split()))
    if n < 3:
        print("NO")
        continue
    left_min = [p[0]]
    for i in range(1, n):
        left_min.append(min(left_min[-1], p[i]))
    right_min = [p[-1]]
    for i in reversed(range(n-1)):
        right_min.append(min(right_min[-1], p[i]))
    # find valid j
    j = -1
    for i in range(1, n-1):
        if left_min[i-1] < p[i] and right_min[n-(i+1)-1] < p[i]:
            j = i
    if j == -1:
        print("NO")
        continue
    print("YES")
    i_ans = 0
    for i in range(j):
        if p[i] < p[j]:
            i_ans = i
            break
    k_ans = 0
    for k in range(j+1, n):
        if p[k] < p[j]:
            k_ans = k
            break
    print(f"{i_ans +1} {j+1} {k_ans +1}")
    
