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

from collections import Counter,defaultdict,deque
#from heapq import *
#from itertools import *
#from operator import itemgetter
#from itertools import count, islice
#from functools import reduce
#alph = 'abcdefghijklmnopqrstuvwxyz'
#dirs = [[1,0],[0,1],[-1,0],[0,-1]]
#from math import factorial as fact
#a,b = [int(x) for x in input().split()]
#sarr = [x for x in input().strip().split()]
#import math
#from math import *

import sys
input=sys.stdin.readline
#sys.setrecursionlimit(2**30)
#MOD = 10**9+7

def solve():
    n = int(input())
    arr = [int(x) for x in input().split()]
    for i in range(1,n-1):
        if arr[i-1]<arr[i]>arr[i+1]:
            print('yes')
            print(i,i+1,i+2)
            return
    print('no')

            
tt = int(input())
for test in range(tt):
    solve()

 


 
 
 
 
 
 
 
 
 
 
 
 
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