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
//#pragma GCC optimize "trapv"
#pragma GCC optimize("Ofast")
#pragma GCC optimize("O3", "unroll-loops")
#pragma GCC target("avx,avx2,fma")
#include<bits/stdc++.h>
#define ll long long int
#define fab(a,b,i) for(ll i=a;i<b;i++)
#define pb push_back
#define db double
#define mp make_pair
//#define endl "\n"
#define f first
#define se second
#define all(x) x.begin(),x.end()
#define MOD 1000000007
#define vll vector<ll>
#define vi vector<int>
#define pii pair<int,int>
#define pll pair<ll,ll>
#define quick ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
ll add(ll x, ll y) {ll res = x + y; return (res >= MOD ? res - MOD : res);}
ll mul(ll x, ll y) {ll res = x * y; return (res >= MOD ? res % MOD : res);}
ll sub(ll x, ll y) {ll res = x - y; return (res < 0 ? res + MOD : res);}
ll power(ll x, ll y) {ll res = 1; x %= MOD; while (y) {if (y & 1)res = mul(res, x); y >>= 1; x = mul(x, x);} return res;}
ll mod_inv(ll x) {return power(x, MOD - 2);}

using namespace std;
const int N=2e5+2;

ll a[N];
ll seg[N*4];
void build(int node,int l, int r)
{
	if(l==r)
	{
		seg[node]=a[l];
	}
	else
	{
		int mid=(l+r)/2;
		build(2*node,l,mid);
		build(2*node+1,mid+1,r);
		seg[node]=min(seg[2*node],seg[2*node+1]);
	}
}
ll query(int node, int l, int r, int s, int e)
{
	//cout<<"s:"<<s<<" "<<e<<endl;
	if(l>e or r<s)
	{
		return 1e9+124;
	}
	if(s<=l and r<=e)
		return seg[node];
	int mid=(l+r)/2;
	ll ans=min(query(2*node,l,mid,s,e),query(2*node+1,mid+1,r,s,e));
	return ans;

}

int main()
{ quick;
#ifndef ONLINE_JUDGE
    freopen("D:/sublime/input.txt", "r", stdin);
    freopen("D:/sublime/output.txt", "w", stdout);
#endif
	int t;
	cin>>t;
	while(t--)
	{
		ll n;
		cin>>n;
		fab(0,n,i)
		{
			cin>>a[i];

		}
		fab(0,4*n+5,i)
		{
			seg[i]=INT_MAX;

		}
		build(1,0,n-1);

		ll pref[n],suff[n];
		pref[0]=a[0];

		map<int,int > m1,m2;

		fab(1,n,i)
		{
			pref[i]=max(pref[i-1],a[i]);
			
		}
		suff[n-1]=a[n-1];
		vector<ll> op={a[n-1]};
		for( int i=n-2;i>=0;i--)
		{
			suff[i]=max(suff[i+1],a[i]);
			
		
		}
		fab(0,n,i)
		{
			m1[suff[i]]=i;

		}
		for( int i=n-1;i>=0;i--)
		{
			m2[suff[i]]=i;
		}
		//reverse(all(op));
		
		 
		vector<int> v;
		fab(0,n-2,i)
		{
		
			if(m1.find(pref[i])==m1.end())
				continue;
			ll get=m2[pref[i]]-1;
			ll ind=m1[pref[i]]-1;
			//cout<<get<<endl;
			

			get=max(i+1,get);

			//cout<<"gg:"<<pref[i]<<" "<<get<<" "<<ind<<" "<<i<<endl;
			while(get<=ind )
			{
				int mid=(get+ind)/2;

				//cout<<"mid:"<<mid<<" "<<10<<endl;
				ll q=query(1,0,n-1,i+1,mid);
				//cout<<"mid:"<<mid<<" "<<q<<endl;
				if(q==pref[i])
				{
					v.pb(i+1);
					v.pb(mid-i);
					v.pb(n-mid-1);
					break;
				}
				else if(q<pref[i])
						ind=mid-1;
				else
					get=mid+1;

			}
			if(v.size())
				break;

		}
		if(v.size()==0)
		{
			cout<<"NO"<<endl;

		}
		else
		{
			cout<<"YES"<<endl;
			for(auto i: v)
				cout<<i<<" ";
			cout<<endl;

		}





	}
	

 cerr << "time taken : " << (float)clock() / CLOCKS_PER_SEC << " secs" << endl;
	return 0;
}