/**
    __author__ : FireGhost
    problems_ID: 1454F
*/

#include <bits/stdc++.h>

using namespace std;

#define mp 					make_pair
#define pb 					push_back
#define fi 					first
#define se 					second
#define all(x)				        x.begin(), x.end()
#define endl				        "\n"
#define _itr				        ::iterator

typedef long long       	                ll;
typedef unsigned long long 	                ull;
typedef vector<int> 		                vi;
typedef pair<int, int>		                pii;

const int N = 1e5 + 3;
const int INF = INT_MAX;
const int NEINF = INT_MIN;
const int MOD = 1e9 + 7;

void Solve()
{
    /// Reading the input
    int n; cin >> n;
    vector <int> a(n + 1);
    for (int i = 1; i <= n; ++i) cin >> a[i];
    /// Maintaining 2 arrays that keep the longest subarray, where a[i] is the
    /// minimum of [min_left[i], min_right[i]].
    /// Also maintaining maximum prefix and maximum suffix arrays.
    deque <int> mi;
    vector <int> min_left(n + 1), min_right(n + 1), pref(n + 1), suff(n + 1);
    pref[1] = a[1]; suff[n] = a[n];
    for (int i = 1; i <= n; ++i)
    {
        while (mi.size() && a[mi.back()] >= a[i]) mi.pop_back();
        if (mi.size() == 0) min_left[i] = 1;
        else min_left[i] = mi.back() + 1;
        mi.push_back(i);
    }
    for (int i = 2; i <= n; ++i)
        pref[i] = max(pref[i - 1], a[i]);
    mi.clear();
    for (int i = n; i >= 1; --i)
    {
        while (mi.size() && a[mi.back()] >= a[i]) mi.pop_back();
        if (mi.size() == 0) min_right[i] = n;
        else min_right[i] = mi.back() - 1;
        mi.push_back(i);
    }
    for (int i = n - 1; i >= 1; --i)
        suff[i] = max(suff[i + 1], a[i]);
    /// Binary search for x and y
    for (int i = 1; i <= n; ++i)
    {
        int x = -1, y = -1;
        int v = a[i];
        int lo = min_left[i] - 1, hi = i - 1;
        while (lo <= hi)
        {
            int mid = (lo + hi) >> 1;
            if (pref[mid] == v)
            {
                x = mid;
                lo = mid + 1;
            }
            else if (pref[mid] > v) hi = mid - 1;
            else lo = mid + 1;
        }
        lo = i + 1, hi = min_right[i] + 1;
        while (lo <= hi)
        {
            int mid = (lo + hi) >> 1;
            if (suff[mid] == v)
            {
                y = mid;
                hi = mid - 1;
            }
            else if (suff[mid] > v) lo = mid + 1;
            else hi = mid - 1;
        }
        if (x == -1 || y == -1) continue;
        cout << "YES" << "\n";
        cout << x << " " << y - x - 1 << " " << n - y + 1 << "\n";
        return;
    }
    cout << "NO" << "\n";
}

int main()
{
    cin.tie(NULL);
    ios_base::sync_with_stdio(false);
    int t; cin >> t;
    while (t--)
        Solve();
    return 0;
}

### Prompt

Your task is to create a CPP solution to the following 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

### Response

```cpp
/**
    __author__ : FireGhost
    problems_ID: 1454F
*/

#include <bits/stdc++.h>

using namespace std;

#define mp 					make_pair
#define pb 					push_back
#define fi 					first
#define se 					second
#define all(x)				        x.begin(), x.end()
#define endl				        "\n"
#define _itr				        ::iterator

typedef long long       	                ll;
typedef unsigned long long 	                ull;
typedef vector<int> 		                vi;
typedef pair<int, int>		                pii;

const int N = 1e5 + 3;
const int INF = INT_MAX;
const int NEINF = INT_MIN;
const int MOD = 1e9 + 7;

void Solve()
{
    /// Reading the input
    int n; cin >> n;
    vector <int> a(n + 1);
    for (int i = 1; i <= n; ++i) cin >> a[i];
    /// Maintaining 2 arrays that keep the longest subarray, where a[i] is the
    /// minimum of [min_left[i], min_right[i]].
    /// Also maintaining maximum prefix and maximum suffix arrays.
    deque <int> mi;
    vector <int> min_left(n + 1), min_right(n + 1), pref(n + 1), suff(n + 1);
    pref[1] = a[1]; suff[n] = a[n];
    for (int i = 1; i <= n; ++i)
    {
        while (mi.size() && a[mi.back()] >= a[i]) mi.pop_back();
        if (mi.size() == 0) min_left[i] = 1;
        else min_left[i] = mi.back() + 1;
        mi.push_back(i);
    }
    for (int i = 2; i <= n; ++i)
        pref[i] = max(pref[i - 1], a[i]);
    mi.clear();
    for (int i = n; i >= 1; --i)
    {
        while (mi.size() && a[mi.back()] >= a[i]) mi.pop_back();
        if (mi.size() == 0) min_right[i] = n;
        else min_right[i] = mi.back() - 1;
        mi.push_back(i);
    }
    for (int i = n - 1; i >= 1; --i)
        suff[i] = max(suff[i + 1], a[i]);
    /// Binary search for x and y
    for (int i = 1; i <= n; ++i)
    {
        int x = -1, y = -1;
        int v = a[i];
        int lo = min_left[i] - 1, hi = i - 1;
        while (lo <= hi)
        {
            int mid = (lo + hi) >> 1;
            if (pref[mid] == v)
            {
                x = mid;
                lo = mid + 1;
            }
            else if (pref[mid] > v) hi = mid - 1;
            else lo = mid + 1;
        }
        lo = i + 1, hi = min_right[i] + 1;
        while (lo <= hi)
        {
            int mid = (lo + hi) >> 1;
            if (suff[mid] == v)
            {
                y = mid;
                hi = mid - 1;
            }
            else if (suff[mid] > v) lo = mid + 1;
            else hi = mid - 1;
        }
        if (x == -1 || y == -1) continue;
        cout << "YES" << "\n";
        cout << x << " " << y - x - 1 << " " << n - y + 1 << "\n";
        return;
    }
    cout << "NO" << "\n";
}

int main()
{
    cin.tie(NULL);
    ios_base::sync_with_stdio(false);
    int t; cin >> t;
    while (t--)
        Solve();
    return 0;
}

```