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 CPP):

Solution:
#include<bits/stdc++.h>
#define int long long int
const int M = 998244353;
const int mod = 1e9+7; 
const int inf = 0;
using namespace std;
#define debug() for(int l=0;l<50;l++){}
#define fast cin.tie(0);cout.tie(0);ios_base::sync_with_stdio(0);
int bs(int arr[], int l, int r, int x){
	int mid=0;
    if (r >= l){
		if (arr[mid] == x) return mid; 
        if (arr[mid] > x) return bs(arr, l, mid - 1, x); 
        return bs(arr, mid + 1, r, x); 
    }
    return -1;
}
int power(int x,int y,int p){
	int res = 1;
	x = x % p;
	if(x == 0) return 0;
	while (y > 0){
		if (y & 1) res = (res*x) % p;
		y = y>>1;
		x = (x*x) % p;
		}
	return res;
}
const int max2=500001;
pair<int,int>st[4*max2];
int arr[max2],arr1[max2];
void buildtree(int si,int ss,int se){
	if(ss==se){
        st[si].first=arr1[ss];
        st[si].second=arr1[ss];
		return;
	}
	int mid=ss+(se-ss)/2;
	buildtree(2*si,ss,mid);
	buildtree(2*si+1,mid+1,se);
	if(arr1[mid]>arr1[mid+1]) st[si].first=st[2*si].first+st[2*si+1].first;
    else st[si].second=st[2*si].second+st[2*si+1].second;
	
}
int query(int si,int ss,int se,int qs,int qe){
	if(qs>se || qe<ss) //completely outside
		return inf;
	if(ss>=qs && se<=qe)
		return st[si].first;
	int mid=ss+(se-ss)/2;
	int l=query(2*si,ss,mid,qs,qe);
	int r=query(2*si+1,mid+1,se,qs,qe);
	return min(l,r);
}
int geteven(int si,int ss,int se,int qs,int qe){
	if(ss>qe || se<qs) //completely outside
		return 0;
	if(ss>=qs && se<=qe)
		return st[si].second;
	int mid=ss+(se-ss)/2;
	int l=geteven(2*si,ss,mid,qs,qe);
	int r=geteven(2*si+1,mid+1,se,qs,qe);
	return l+r;
}
int getodd(int si,int ss,int se,int qs,int qe){
	if(qs>se || qe<ss) //completely outside
		return 0;
	if(ss>=qs && se<=qe)
		return st[si].first;
	int mid=ss+(se-ss)/2;
	int l=getodd(2*si,ss,mid,qs,qe);
	int r=getodd(2*si+1,mid+1,se,qs,qe);
	return l+r;
}

void update(int si,int ss,int se,int qi,int newval){
	if(ss==se){
        if(arr[ss]%2==1) st[si]={0,1};
        else st[si]={1,0};
		arr[ss]=newval;
		return;
	}
	int mid=ss+(se-ss)/2;
	if(qi<=mid) update(2*si,ss,mid,qi,newval);
	else update(2*si+1,mid+1,se,qi,newval);
	st[si].first=st[2*si].first+st[2*si+1].first;
    st[si].second=st[2*si].second+st[2*si+1].second;
}

int mm(int a, int b, int m){
    int res = 0;
    a %= m; 
    while (b){  
        if (b & 1) 
            res = (res + a) % m; 
        a = (2 * a) % m; 
        b >>= 1; 
    }
    return res; 
} 
/*int dp[100000];
void sieve(int n){ 
    bool prime[n+1]; 
    memset(prime, true, sizeof(prime)); 
    for (int p=2; p*p<=n; p++){
        if (prime[p] == true){
            for (int i=p*p; i<=n; i += p) 
                prime[i] = false; 
        } 
    } 
    int i=0;
    for (int p=2; p<=n; p++){ 
       if (prime[p]){
		   dp[i]=p;
		   i++;
	   }
   }
} */
int dp[200001][32];
void early(int a[],int n){
	for(int i=0;i<n;i++) dp[i][0]=i;
	for(int j=1;(1<<j)<=n;j++){
		for(int i=0;(i+(1<<j)-1)<n;i++){
			if(a[dp[i][j-1]]<a[dp[i+(1<<(j-1))][j-1]])
				dp[i][j]=dp[i][j-1];
			else dp[i][j]=dp[i+(1<<(j-1))][j-1];
		}
	}
}
int rmq(int a[],int l,int r){
	int j=(int)log2(r-l+1);
	if(a[dp[l][j]]<=a[dp[r-(1<<j)+1][j]])
		return a[dp[l][j]];
	else return a[dp[r-(1<<j)+1][j]];
}
void answer(int idx1,int idx2,int n)
{
	cout<<"YES\n"<<idx1+1<<' '<<idx2-idx1<<' '<<n-idx2-1<<'\n';
	return;
}	
void solve(){
	int n;
	cin>>n;
	int a[n],pre[n],post[n];
	map<int,int>m;
	for(int i=0;i<n;i++){ cin>>a[i]; m[a[i]]=i;}
	pre[0]=a[0]; post[n-1]=a[n-1];
	for(int i=1;i<n;i++) pre[i]=max(pre[i-1],a[i]);
	for(int i=n-2;i>=0;i--) post[i]=max(post[i+1],a[i]);
	early(a,n);
	
	for(int i=0;i<n-2;i++)
	{
		int max1=pre[i];
		int l=i+1,r=m[max1];
		while(l<=r)
		{
			int mid=(l+r)/2;
			if(rmq(a,i+1,mid)>=max1)
			{
				if(post[mid+1]==max1 && rmq(a,i+1,mid)==max1)
				{
					answer(i,mid,n);
					return;
				}
				l=mid+1;
			}
			else r=mid-1;
		}
	}
	cout<<"NO\n";
	
}
signed main(){
	fast;
	int t = 1;
	cin>>t;
	while(t--) solve();
	return 0;
}