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 deque
from collections import OrderedDict
import math
 
import sys
import os
import threading
import bisect

import operator
 
import heapq


from atexit import register
from io import BytesIO

#sys.stdin = BytesIO(os.read(0, os.fstat(0).st_size))
#sys.stdout = BytesIO()
#register(lambda: os.write(1, sys.stdout.getvalue()))


import io
#input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
#sys.stdin = open("F:\PY\\test.txt", "r")
input = lambda: sys.stdin.readline().rstrip("\r\n")
#input = sys.stdin.readline


for t in range(int(input())):
    n = int(input())
    arr = [0]+list(map(int, input().split()))
    dpl = [1005]*1005
    dpr = [1005]*1005
    #iL = [0]*1005
    #iR = [0]*1005
    for i in range(1, n+1):
        dpl[i]=min(dpl[i-1], arr[i])
    for i in range(n, 0, -1):
        dpr[i]=min(dpr[i+1], arr[i])        
    for i in range(1, n+1):
        if dpl[i]<arr[i] and dpr[i]<arr[i]:
            print("YES")
            for j in range(i):
                if arr[j]==dpl[i]:
                    print(j, end=" ")
            print(i, end=" ")
            for j in range(i+1, n+1):
                if arr[j]==dpr[i]:
                    print(j)  
            break
    else:
        print("NO")
            
            









