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
inputlist=sys.stdin.readlines()
number_of_testcases=int(inputlist[0].strip())
i=0
while(i<2*number_of_testcases):
    #print(i)
    i+=1
    n=int(inputlist[i].strip())
    i+=1
    numbers_list=list(map(int,inputlist[i].strip().split(' ')))
    min1=numbers_list[0]
    min1_index=0
    max_now=-1
    max_index=-1
    for j in range(1,n):
        if(numbers_list[j]>min1):
            max_now=numbers_list[j]
            max_index=j
            break
        else:
            min1=numbers_list[j]
            min1_index=j
    else:
        print('NO')
        continue
    for k in range(j+1,n):
        if(numbers_list[k]<max_now):
            min2_index=k
            print('YES')
            print(min1_index+1,max_index+1,min2_index+1)
            break
        else:
            max_now=numbers_list[k]
            max_index=k
    else:
        print('NO')