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())

while (t):

	n = int(input())
	a = input().split()
	a = list(map(int,a))

	m = [0 for i in range(n)]
	m[-1] = n-1
	for i in range(n-2,-1,-1):
		if (a[i] < a[m[i+1]]):
			m[i] = i
		else:
			m[i] = m[i+1]

	x,y,z = 0,0,0
	for i in range(n-2):
		for j in range(i+1,n-1):
			if (a[i] < a[j] and a[j] > a[m[j+1]]):
				x,y,z = i+1,j+1,m[j+1]+1
			if (x):
				break
		if (x):
			break

	if (x):
		print('YES')
		print(x,y,z)
	else:
		print('NO')

	t -= 1
		