Problem: 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

Solution (in PYTHON3):

Solution:
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
    newlines = 0
 
    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None
 
    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()
 
    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)
 
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")
 
 
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline()
 
# ------------------------------

def RL(): return map(int, sys.stdin.readline().split())
def RLL(): return list(map(int, sys.stdin.readline().split()))
def N(): return int(input())
def print_list(l):
    print(' '.join(map(str,l)))
    
# sys.setrecursionlimit(100000)
# import random
# from functools import reduce
# from functools import lru_cache
# from heapq import *
# from collections import deque as dq
# from math import ceil,floor,sqrt,pow,gcd,log
# import bisect as bs
# from collections import Counter
# from collections import defaultdict as dc 

def sparse_table(a):
    n = len(a)
    table = [a]
    k = 0
    while n > 1 << k + 1:
        this = [0]*(n - (1 << k + 1) + 1)
        for i in range(n - (1 << k + 1) + 1):
            this[i] = min(table[-1][i],table[-1][i+(1<<k)])
        table.append(this)
        k+=1
    return table
def get_range_minimal(table,i,j):
    k, t = 0, (j - i - 1) >> 1
    while t:
        t >>= 1
        k += 1
    return min(table[k][i],table[k][j - (1 << k)])

for _ in range(N()):
    n = N()
    a = RLL()
    m = max(a)
    if a.count(m) > 2:
        key = a.index(m)
        key = a.index(m, key + 1)
        print('YES')
        print(key, 1, n - key - 1)
    else:
        table = sparse_table(a)
        flag = False
        l, r, ml, mr = 0, n, 0, 0
        while ml < m:
            if a[l] == ml:
                ml = a[l + 1]
                l += 2
            else:
                ml = a[l]
                l += 1
            if a[r - 1] == mr:
                mr = a[r - 2]
                r -= 2
            else:
                mr = a[r - 1]
                r -= 1
            while ml != mr:
                while ml < mr:
                    ml = max(ml, a[l])
                    l += 1
                while mr < ml:
                    r -= 1
                    mr = max(mr, a[r])
            if ml == m:
                break
            for i in range(l, r):
                if a[i] > ml:
                    break
                l = i + (a[i] < ml)
            for i in range(r - 1, l - 1, -1):
                if a[i] > mr:
                    break
                r = i + (a[i] == mr)
            if l < r and get_range_minimal(table, l, r) == ml:
                print('YES')
                print(l, r - l, n - r)
                flag = True
                break
        if not flag:
            print('NO')