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 sys, math
import io, os
#data = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
from bisect import bisect_left as bl, bisect_right as br, insort
from heapq import heapify, heappush, heappop
from collections import defaultdict as dd, deque, Counter
from itertools import permutations,combinations
def data(): return sys.stdin.readline().strip()
def mdata(): return list(map(int, data().split()))
def outl(var) : sys.stdout.write(' '.join(map(str, var))+'\n')
def out(var) : sys.stdout.write(str(var)+'\n')
#from decimal import Decimal
#from fractions import Fraction
#sys.setrecursionlimit(100000)
INF = float('inf')
mod = 998244353


class SegmentTree:
    def __init__(self, data, default=INF, func=min):
        self._default = default
        self._func = func
        self._len = len(data)
        self._size = _size = 1 << (self._len - 1).bit_length()

        self.data = [default] * (2 * _size)
        self.data[_size:_size + self._len] = data
        for i in reversed(range(_size)):
            self.data[i] = func(self.data[i + i], self.data[i + i + 1])

    def __delitem__(self, idx):
        self[idx] = self._default

    def __getitem__(self, idx):
        return self.data[idx + self._size]

    def __setitem__(self, idx, value):
        idx += self._size
        self.data[idx] = value
        while idx:
            idx >>= 1
            self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])

    def __len__(self):
        return self._len

    def query(self, start, stop):
        """func of data[start, stop)"""
        start += self._size
        stop += self._size

        res_left = res_right = self._default
        while start < stop:
            if start & 1:
                res_left = self._func(res_left, self.data[start])
                start += 1
            if stop & 1:
                stop -= 1
                res_right = self._func(self.data[stop], res_right)
            start >>= 1
            stop >>= 1

        return self._func(res_left, res_right)

    def __repr__(self):
        return "SegmentTree({0})".format(self.data)


for t in range(int(data())):
    n=int(data())
    a=mdata()
    S=SegmentTree(a)
    m=[[a[-1],1]]
    d=dd(int)
    d[a[-1]]=n-1
    for i in range(n-2,-1,-1):
        if a[i]<=m[-1][0]:
            m.append(m[-1])
            if a[i]==m[-1][0]:
                m[-1][1]+=1
            d[m[-1][0]]=i
        else:
            m.append([a[i],1])
            d[a[i]]=i
    m=m[::-1]
    m1=a[0]
    flag=False
    l=[]
    if m[0][1]>2:
        flag=True
        l=[]
        for i in range(n):
            if len(l)==3:
                break
            if a[i]==m[0][0]:
                l.append(i+1)
        l[0],l[1],l[2]=l[1]-1,1,n-l[1]
    else:
        for i in range(n):
            m1=max(m1,a[i])
            if m1 in d and d[m1]>i:
                m2=S.query(i+1,d[m1])
                if m2==m1:
                    flag=True
                    l=[i+1,d[m1]-i-1,n-d[m1]]
                elif m2>m1 and a[d[m1]]==m1 and m[d[m1]][1]>1:
                    l = [i + 1, d[m1] - i, n - d[m1]-1]
                    flag=True
            if flag:
                break
    if flag:
        out("YES")
        outl(l)
    else:
        out("NO")
