You are given an array a consisting of n integers.

Let min(l, r) be the minimum value among a_l, a_{l + 1}, …, a_r and max(l, r) be the maximum value among a_l, a_{l + 1}, …, a_r.

Your task is to choose three positive (greater than 0) integers x, y and z such that:

  * x + y + z = n; 
  * max(1, x) = min(x + 1, x + y) = max(x + y + 1, n). 



In other words, you have to split the array a into three consecutive non-empty parts that cover the whole array and the maximum in the first part equals the minimum in the second part and equals the maximum in the third part (or determine it is impossible to find such a partition).

Among all such triples (partitions), you can choose any.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.

The first line of the test case contains one integer n (3 ≤ n ≤ 2 ⋅ 10^5) — the length of a.

The second line of the test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.

It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).

Output

For each test case, print the answer: NO in the only line if there is no such partition of a that satisfies the conditions from the problem statement. Otherwise, print YES in the first line and three integers x, y and z (x + y + z = n) in the second line.

If there are several answers, you can print any.

Example

Input


6
11
1 2 3 3 3 4 4 3 4 2 1
8
2 9 1 7 3 9 4 1
9
2 1 4 2 4 3 3 1 2
7
4 2 1 1 4 1 4
5
1 1 1 1 1
7
4 3 4 3 3 3 4


Output


YES
6 1 4
NO
YES
2 5 2
YES
4 1 2
YES
1 1 3
YES
2 1 4
import io,os
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
import heapq

 
T = int(input())
r = 1
 
 
while r<=T:
    n = int(input())
    arr = list(map(int,input().split()))


#    arr = [1,2]*50000+[1,3,1,2]+[1,2]*49998
#    n = 200000

    third = [0]*n
    third[-1] = arr[-1]
    maxfre = 0
    for i in range(n-2,-1,-1):
        third[i] = max(third[i+1],arr[i])

    firstdic = {}
     
 

    temp = []
    secondindexnum = 0
    secondindexmin = 2147483647

    flag = False

    fuck = 0
 
    for i in range(n-1):
        if arr[i]<=third[i+1]: 
            heapq.heappush(temp,[-arr[i],-i])
            if arr[i] not in firstdic:  firstdic[arr[i]] = []
            firstdic[arr[i]].append(i)
        else:
            secondindexnum += 1
            secondindexmin = min(secondindexmin,i)



        while temp and -temp[0][0]>third[i+1]:
            ele = heapq.heappop(temp)
            secondindexnum += 1
            secondindexmin = min(secondindexmin,-ele[1])
            if -ele[0] in firstdic:  del firstdic[-ele[0]]



 
        if len(temp)<2: continue
        if temp[0][0]!=-third[i+1]: continue
        maximum = heapq.heappop(temp)
        if temp[0][0]!=-third[i+1]:
            heapq.heappush(temp,maximum)
            continue    
        heapq.heappush(temp,maximum)   

 
        if secondindexnum==0:
            flag = True
            x,y,z = len(temp)-1,1,n-len(temp)
            break   
        needed = i - secondindexmin + 1 - secondindexnum
        if len(temp)<needed+1: continue


#        print("hahaha",needed)
 

        if needed==0 and -temp[0][1]==secondindexmin-1:
            flag = True
            x,y,z = secondindexmin-1,i+2-secondindexmin,n-i-1
            break


        if needed>0 and len(firstdic[third[i+1]])>needed:

            target = firstdic[third[i+1]] 
            if target[-needed]>secondindexmin:
                flag = True
                x,y,z = secondindexmin,i+1-secondindexmin,n-i-1


            if flag: break


            
   

 
 
        
 
        
    if flag:
        print("YES")
        print(x,y,z) 
    else:
        print("NO")       
 
        
 
        
    
 
 
 
 
 
 
 
 
     
 
 
 
    r += 1
