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")
#pragma GCC target ("sse4")

#include<iostream>
#include<string>
#include<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<bitset>
#include<stack>
#include<unordered_map>
#include<unordered_set>
#include<utility>
#include<cassert>
#include<complex>
#include<numeric>
#include<array>
using namespace std;

//#define int long long
typedef long long ll;

typedef unsigned long long ul;
typedef unsigned int ui;
constexpr ll mod = 998244353;
const ll INF = mod * mod;
typedef pair<int, int>P;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
#define all(v) (v).begin(),(v).end()
typedef pair<ll, ll> LP;
typedef double ld;
typedef pair<ld, ld> LDP;
const ld eps = 1e-12;
const ld pi = acosl(-1.0);

ll mod_pow(ll x, ll n, ll m = mod) {
	ll res = 1;
	while (n) {
		if (n & 1)res = res * x % m;
		x = x * x % m; n >>= 1;
	}
	return res;
}
struct modint {
	ll n;
	modint() :n(0) { ; }
	modint(ll m) :n(m) {
		if (n >= mod)n %= mod;
		else if (n < 0)n = (n % mod + mod) % mod;
	}
	operator int() { return n; }
};
bool operator==(modint a, modint b) { return a.n == b.n; }
modint operator+=(modint& a, modint b) { a.n += b.n; if (a.n >= mod)a.n -= mod; return a; }
modint operator-=(modint& a, modint b) { a.n -= b.n; if (a.n < 0)a.n += mod; return a; }
modint operator*=(modint& a, modint b) { a.n = ((ll)a.n * b.n) % mod; return a; }
modint operator+(modint a, modint b) { return a += b; }
modint operator-(modint a, modint b) { return a -= b; }
modint operator*(modint a, modint b) { return a *= b; }
modint operator^(modint a, ll n) {
	if (n == 0)return modint(1);
	modint res = (a * a) ^ (n / 2);
	if (n % 2)res = res * a;
	return res;
}

ll inv(ll a, ll p) {
	return (a == 1 ? 1 : (1 - p * inv(p % a, a)) / a + p);
}
modint operator/(modint a, modint b) { return a * modint(inv(b, mod)); }

const int max_n = 1 << 18;
modint fact[max_n], factinv[max_n];
void init_f() {
	fact[0] = modint(1);
	for (int i = 0; i < max_n - 1; i++) {
		fact[i + 1] = fact[i] * modint(i + 1);
	}
	factinv[max_n - 1] = modint(1) / fact[max_n - 1];
	for (int i = max_n - 2; i >= 0; i--) {
		factinv[i] = factinv[i + 1] * modint(i + 1);
	}
}
modint comb(int a, int b) {
	if (a < 0 || b < 0 || a < b)return 0;
	return fact[a] * factinv[b] * factinv[a - b];
}


struct sparse_table {
private:
	int n;
	int tmp;
	vector<int> ori;
	vector<vector<int>> nodel, noder;
public:
	sparse_table(vector<int> _ori) {
		ori = _ori;
		n = ori.size();
		int cop = n;
		tmp = 0;
		while (cop > 0) {
			tmp++; cop >>= 1;
		}
		nodel.resize(n);
		noder.resize(n);
		rep(i, n) {
			nodel[i].resize(tmp);
			noder[i].resize(tmp);
		}
		rep(i, n) {
			nodel[i][0] = ori[i];
			noder[i][0] = ori[i];
		}
		rep(j, tmp - 1) {
			rep(i, n) {
				if (i + (1 << j) < n) {
					nodel[i][j + 1] = min(nodel[i][j], nodel[i + (1 << j)][j]);
				}
				if (i - (1 << j) >= 0) {
					noder[i][j + 1] = min(noder[i][j], noder[i - (1 << j)][j]);
				}
			}
		}
	}
	//[l,r)
	int query(int l, int r) {
		if (r - l == 0)return mod;
		int len = r - l;
		int x = 1;
		int cnt = 0;
		while (x * 2 <= len) {
			x *= 2;
			cnt++;
		}
		return min(nodel[l][cnt], noder[r - 1][cnt]);
	}
};

void solve() {
	int n; cin >> n;
	vector<int> a(n);
	rep(i, n)cin >> a[i];
	sparse_table st(a);

	int x = a[n - 1];
	vector<int> r;
	per(i, n) {
		x = max(x, a[i]);
		r.push_back(x);
	}
	int ma = a[0];
	//[0,i]
	rep(i, n-2) {
		ma = max(ma, a[i]);
		int le = i+1, ri = n+1;
		//[i+1,mid)
		while (ri - le > 1) {
			int mid = (le + ri) / 2;
			int val = st.query(i + 1, mid);
			if (val < ma)ri = mid;
			else le = mid;
		}
		int r1 = le+1;
		le = i+1, ri = n+1;
		while (ri - le > 1) {
			int mid = (le + ri) / 2;
			int val = st.query(i + 1, mid);
			if (val <= ma)ri = mid;
			else le = mid;
		}
		int l1 = le+1;
		int r2 = lower_bound(all(r), ma) - r.begin();
		int l2 = upper_bound(all(r), ma) - r.begin();
		l2 = n - l2;
		r2 = n - r2;

		//cout << i << " " << l1 << " " << l2 << " " << r1 << " " << r2 << "\n";
		l1 = max(l1, l2);
		r1 = min(r1, r2);
		if (l1 < r1) {
			cout << "YES\n";
			cout << i + 1 << " " << l1 - (i+1) << " " << n - l1 << "\n";
			return;
		}
	}
	cout << "NO\n";
}

signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	//cout << fixed << setprecision(15);
	//init_f();
	//init();
	//expr();
	int t; cin >> t; rep(i, t)
	solve();
	return 0;
}