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:
#pragma GCC optimize("Ofast")

#include <bits/stdc++.h>
using namespace std;

#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define pop_b pop_back
#define pf push_front
#define pop_f pop_front
#define mt make_tuple
#define eb emplace_back
#define sf(x) sizeof(x)
#define len(x) x.length()
#define sz(x) x.size()
#define fastio() ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)

#define LB(x,v) (lower_bound(x.begin(),x.end(),v)-x.begin())
#define UB(x,v) (upper_bound(x.begin(),x.end(),v)-x.begin())
typedef long long ll;
typedef pair<ll,ll> ii;
typedef long double ld; 
typedef vector<int> vi;
typedef vector<set<ll>> vset;
typedef vector<ll> vl;
typedef vector<ld> vd;
typedef deque<ll> dql;
typedef deque<int> dqi;
typedef unsigned long long ull;
typedef tuple<int,int,int> State;

#define error(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args); }
void err(istream_iterator<string> it) {}
template<typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
	cerr << *it << " = " << a << "\n";
	err(++it, args...);
}

#define vmax(type)   std::numeric_limits<type>::max()
#define vmin(type)   std::numeric_limits<type>::min()
#define each(a,x) for( auto &a:(x))
#define rep(i, n) for (ll i = 0; i < (n); i++)
#define repr(i, n) for (ll i = (n) - 1; i >= 0; i--)
#define rep2(i, l, r) for (ll i = (l); i < (r); i++)
#define rep2r(i, l, r) for (ll i = (r) - 1; i >= (l); i--)
#define range(a) a.begin(), a.end()
#define deb(x) cout<<#x<<" "<<x<<endl;
#define cinai(a, n) vi a(n); rep(i, n) { cin>>a[i]; }
#define cinal(a, n) vl a(n); rep(i, n) { cin>>a[i]; }
#define coutai(a, n) rep(i, n) { cout<<a[i]<<" "; } ent;
#define coutal(a, n) rep(i, n) { cout<<a[i]<<" "; } ent;
#define ent cout<<endl;
#define resize_vec(a) a.resize(unique(a.begin(), a.end()) - a.begin());
#define ms(dp, x) memset(dp, x, sf(dp))
#define endl '\n'

ll powm(ll a, ll b, ll mod) {
	ll res=1;
	while(b) {
		if(b&1)
			res=(res*a)%mod;
		a=(a*a)%mod;
		b>>=1;
	}
	return res;
}

int db = 0;
const ll mod = 1e9 + 7;
const ll N = 1000005;
const ll inf = 1e9;
#define er if(db) error

template<typename T> struct rmq {
	vector<T> v; int n;
	static const int b = 30; // block size
	vector<int> mask, t; // mask and sparse table

	int op(int x, int y) {
		return v[x] < v[y] ? x : y;
	}
	// least significant set bit
	int lsb(int x) {
		return x & -x;
	}
	// index of the most significant set bit
	int msb_index(int x) {
		return __builtin_clz(1)-__builtin_clz(x);
	}
	// answer query of v[r-size+1..r] using the masks, given size <= b
	int small(int r, int size = b) {
		// get only 'size' least significant bits of the mask
		// and then get the index of the msb of that
		int dist_from_r = msb_index(mask[r] & ((1<<size)-1));

		return r - dist_from_r;
	}
	rmq(const vector<T>& v_) : v(v_), n(v.size()), mask(n), t(n) {
		int curr_mask = 0;
		for (int i = 0; i < n; i++) {

			// shift mask by 1, keeping only the 'b' least significant bits
			curr_mask = (curr_mask<<1) & ((1<<b)-1);

			while (curr_mask > 0 and op(i, i - msb_index(lsb(curr_mask))) == i) {
				// current value is smaller than the value represented by the
				// last 1 in curr_mask, so we need to turn off that bit
				curr_mask ^= lsb(curr_mask);
			}
			// append extra 1 to the mask
			curr_mask |= 1;

			mask[i] = curr_mask;
		}

		// build sparse table over the n/b blocks
		// the sparse table is linearized, so what would be at
		// table[j][i] is stored in table[(n/b)*j + i]
		for (int i = 0; i < n/b; i++) t[i] = small(b*i+b-1);
		for (int j = 1; (1<<j) <= n/b; j++) for (int i = 0; i+(1<<j) <= n/b; i++)
			t[n/b*j+i] = op(t[n/b*(j-1)+i], t[n/b*(j-1)+i+(1<<(j-1))]);
	}
	// query(l, r) returns the actual minimum of v[l..r]
	// to get the index, just change the first and last lines of the function
	T query(int l, int r) {
		// query too small
		if (r-l+1 <= b) return v[small(r, r-l+1)];

		// get the minimum of the endpoints
		// (there is no problem if the ranges overlap with the sparse table query)
		int ans = op(small(l+b-1), small(r));

		// 'x' and 'y' are the blocks we need to query over
		int x = l/b+1, y = r/b-1;

		if (x <= y) {
			int j = msb_index(y-x+1);
			ans = op(ans, op(t[n/b*j+x], t[n/b*j+y-(1<<j)+1]));
		}

		return v[ans];
	}
};

void solve()
{
	int n;
	cin>>n;
	cinal(a, n);
	rmq<ll> pre(a);
	ll cur_min = a[0];
	ll rev_max[n];
	rev_max[n-1] = a[n-1];
	repr(i, n-1)
		rev_max[i] = max(rev_max[i+1], a[i]);
	for(int i = 0; i < n - 2; i++){
		cur_min = max(a[i], cur_min);
		int l = i+1, r = n-2;
		while(l < r){
			int mid = (l + r + 1) / 2;
			// cout<<i<<" "<<mid<<" "<<cur_min<<" "<<pre.query(i+1, mid)<<" "<<rev_max[mid+1]<<" "<<l<<" "<<r<<endl;
			if(cur_min == pre.query(i+1, mid)){
				if(rev_max[mid+1] == cur_min){
					cout<<"YES\n";
					cout<<i+1<<" "<<mid - i<<" "<<n - mid - 1<<endl;
					return;
				}else if(rev_max[mid+1] > cur_min){
					l = mid;
				}else
					r = mid - 1;
			}
			else if(cur_min > pre.query(i+1, mid)){
				r = mid - 1;
			}
			else{
				l = mid;
			}
		}
		if(cur_min == pre.query(i+1, l) and rev_max[l+1] == cur_min){
			cout<<"YES\n";
			cout<<i+1<<" "<<l - i<<" "<<n - l - 1<<endl;
			return;
		}
	}
	cout<<"NO\n";
}

// #define _DEBUG
#define testCases
signed main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
	cout << fixed << setprecision(15);
	
	// freopen("input.txt", "r", stdin);
	// freopen("output.txt", "w", stdout);

	int tc = 1;
#ifdef testCases
	cin>>tc;
#endif
	for(int t = 1; t <= tc; t++)
	{
#ifdef _DEBUG
	db = 1;
	int tt = clock();
#endif
		// cout<<"Case #"<<t<<": ";
		solve();
		
#ifdef _DEBUG
		cerr << "TIME = " << clock() - tt << endl;
		tt = clock();
#endif
	}
	return 0;
}