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 sys
import math
from collections import defaultdict,Counter

# input=sys.stdin.readline
# def print(x):
#     sys.stdout.write(str(x)+"\n")

import os
import sys
from io import BytesIO, IOBase

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().rstrip("\r\n")

# sys.stdout=open("CP3/output.txt",'w')
# sys.stdin=open("CP3/input.txt",'r')

from math import ceil,log2

INT_MAX = sys.maxsize

# A utility function to get
# minimum of two numbers
def minVal(x, y) :
	return x if (x < y) else y

# A utility function to get the
# middle index from corner indexes.
def getMid(s, e) :
	return s + (e - s) // 2

def RMQUtil( st, ss, se, qs, qe, index) :

	# If segment of this node is a part
	# of given range, then return
	# the min of the segment
	if (qs <= ss and qe >= se) :
		return st[index]

	# If segment of this node
	# is outside the given range
	if (se < qs or ss > qe) :
		return INT_MAX

	# If a part of this segment
	# overlaps with the given range
	mid = getMid(ss, se)
	return minVal(RMQUtil(st, ss, mid, qs,
						qe, 2 * index + 1),
				RMQUtil(st, mid + 1, se,
						qs, qe, 2 * index + 2))

# Return minimum of elements in range
# from index qs (query start) to
# qe (query end). It mainly uses RMQUtil()
def RMQ( st, n, qs, qe) :

	return RMQUtil(st, 0, n - 1, qs, qe, 0)

# A recursive function that constructs
# Segment Tree for array[ss..se].
# si is index of current node in segment tree st
def constructSTUtil(arr, ss, se, st, si) :

	# If there is one element in array,
	# store it in current node of
	# segment tree and return
	if (ss == se) :

		st[si] = arr[ss]
		return arr[ss]

	# If there are more than one elements,
	# then recur for left and right subtrees
	# and store the minimum of two values in this node
	mid = getMid(ss, se)
	st[si] = minVal(constructSTUtil(arr, ss, mid,
									st, si * 2 + 1),
					constructSTUtil(arr, mid + 1, se,
									st, si * 2 + 2))

	return st[si]

def constructST( arr, n) :

	# Allocate memory for segment tree

	# Height of segment tree
	x = int(ceil(log2(n)))

	# Maximum size of segment tree
	max_size = 2 * 2**x - 1

	st = [0] * max_size

	# Fill the allocated memory st
	constructSTUtil(arr, 0, n - 1, st, 0)

	# Return the constructed segment tree
	return st

# mod=pow(10,9)+7
t=int(input())
for i in range(t):
	n=int(input())
	a=list(map(int,input().split()))
	st = constructST(a, n)
	d=defaultdict(list)
	ma=0
	near=defaultdict(list)
	for j in range(n-1,-1,-1):
		ma=max(ma,a[j])
		d[ma].append(j)
		near[a[j]].append(j)
	# print(near)
	ma=0
	for j in range(n-2):
		ma=max(ma,a[j])
		while near[ma] and near[ma][-1]<=j:
			near[ma].pop()
		if near[ma]:
			while d[ma] and d[ma][-1]<=near[ma][-1]:
				d[ma].pop()
			if d[ma]:
				mi=RMQ(st, n, j+1, d[ma][-1]-1)
				if mi==ma:
					print("YES")
					print(j+1,d[ma][-1]-j-1,n-d[ma][-1])
					break
	else:
		print("NO")