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
def inp():
    return(int(input()))
def inlt():
    return(list(map(int,input().split())))
def insr():
    s = input()
    return(list(s[:len(s) ]))
def invr():
    return(map(int,input().split()))


def solve():
    k = inp()

    l = inlt()

    a = l[0]
    ind_a = 0
    b = 0
    ind_b = 0
    c = 0
    ind_c = 0

    for i in range(k):
        if(l[i] < a and c == 0 and b == 0):
            a = l[i]
            ind_a = i
        elif(l[i] > a and c == 0 and l[i] > b):
            b = l[i]
            ind_b = i

        if(l[i] < b):
            c = l[i]
            ind_c = i
        #print(ind_a, ind_b, ind_c)
    if(ind_c != 0):
        print("YES")
        print(str(ind_a + 1) + " " + str(ind_b + 1) + " " + str(ind_c+1))
    else:
        print("NO")



n = inp()


for _ in range(n):
    solve()
