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
import sys, math
#map(int, sys.stdin.readline().split())
#list(map(int, sys.stdin.readline().split()))

t = int(sys.stdin.readline())

for _ in range(t):
	n = int(sys.stdin.readline())
	liste = list(map(int, sys.stdin.readline().split()))
	ans = "NO"
	out = [0, 0, 0]
	pp = [10001]*n
	pg = [10001]*n
	val = [0, 0, 0]
	for j in range(1, n):
		pp[j] = min(pp[j-1], liste[j-1])
	for j in range(n-2, -1, -1):
		pg[j] = min(pg[j+1], liste[j+1])
		if pg[j] < liste[j] and pp[j] < liste[j]:
			ans = "YES"
			val = [pp[j], liste[j], pg[j]]
	for i in range(n):
		if liste[i] == val[0]:
			out[0] = i+1
		if liste[i] == val[1]:
			out[1] = i+1
		if liste[i] == val[2]:
			out[2] = i+1
	print(ans)
	if ans == "YES":
		print(*out)
	#print(pp, pg)