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
def sol(arr: list)->tuple:
    minl = []
    prev = float('inf')
    for i in arr:
        prev = min(prev,i)
        minl.append(prev)
    minr = []
    prev = float('inf')
    for i in reversed(arr): 
        prev = min(prev,i)
        minr.append(prev)
    minr.reverse()
    for i in range(len(arr)):
        if minl[i]<arr[i] and arr[i]>minr[i]:
            for l in range(i):
                if arr[l]<arr[i]:
                    break
            for r in range(i,len(arr)):
                if arr[r]<arr[i]:
                    break
            return [l,i,r]
    else:
        return []

T = int(input())
tmp = []
p = print
# def print(*args,**kwargs):
#     tmp.append((args,kwargs))
# def flush():
#     for args,kwargs in tmp:
#         p(*args,**kwargs)
#     return

for _ in range(T):
    n = input()
    row = [*map(int,input().split(' '))]
    out = sol(row)
    if out:
        print("YES")
        i,j,k = out
        print(i+1,j+1,k+1)
    else:
        print("NO")
# flush()