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 math 
from collections import deque
import sys


sys.setrecursionlimit(10**4) 

def Divisors(n) : 
	
	l=[]
	i = 2
	while i <= math.sqrt(n): 
		
		if (n % i == 0) : 
			
			
			if (n // i == i) : 
				l.append(i)
			else : 
				l.append(i)
				l.append(n//i)
		i = i + 1
	return l

def SieveOfEratosthenes(n): 
	
	l=[]
	
	prime = [True for i in range(n+1)] 
	p = 2
	while (p * p <= n): 
		
		if (prime[p] == True): 
			
			for i in range(p * p, n+1, p): 
				prime[i] = False
		p += 1
	
	for p in range(2, n+1): 
		if prime[p]: 
			l.append(p)
	
	return l		

def primeFactors(n): 
	
	l=[]
	
	while n % 2 == 0: 
		l.append(2) 
		n = n / 2
		
	for i in range(3,int(math.sqrt(n))+1,2): 
		
		while n % i== 0: 
			l.append(i) 
			n = n / i 
			
	if n > 2: 
		l.append(n)
	
	return(l)	
		

def Factors(n) : 
	
	
	result = []
	
	for i in range(2,(int)(math.sqrt(n))+1) : 

		if (n % i == 0) : 
		    
			if (i == (n/i)) : 
				result.append(i)
			else : 
			    result.append(i)
			    result.append(n//i)
		
	result.append(1)
	
	return result

def maxSubArraySum(a): 
	
	max_so_far = 0
	max_ending_here = 0
	size=len(a)
	
	for i in range(0, size): 
		max_ending_here = max_ending_here + a[i] 
		if (max_so_far < abs(max_ending_here)): 
			max_so_far = max_ending_here 

	return max_so_far 

def longestsubarray(arr, n, k): 
    current_count = 0
      
    # this will contain length of  
    # longest subarray found 
    max_count = 0
  
    for i in range(0, n, 1): 
        if (arr[i] % k != 0): 
            current_count += 1
        else: 
            current_count = 0
        max_count = max(current_count,  
                            max_count) 
      
    return max_count 
    
#print(SieveOfEratosthenes(100))
#print(Divisors(100))
#print(primeFactors(100))
#print(Factors(100))
#print(maxSubArraySum(a))
 

def main():
    
    n=int(input())
    l=list(map(int,input().split()))
    f=0
    for j in range(1,n-1):
        
        if min(l[0:j]) < l[j] and min(l[j+1:n])<l[j]: 
            f=1
            break
        
    if f==1:
        print("YES")
        m=n
        ans1=0
        for k in range(0,j):
            if l[k]<m:
                m=l[k]
                ans1=k
        m=n
        ans2=0
        for k in range(j+1,n):
            if l[k]<m:
                m=l[k]
                ans2=k        
        print(ans1+1,j+1,ans2+1)        
            
    else:
        print("NO")
      
    
          
t=int(input())
for i in range(0,t):
    main()