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 <iostream>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <set>
#include <utility>
#include <vector>
#include <list>
#include <algorithm>
#include <regex>
#include <cmath>
#include <map>
#include <iterator>
#include <climits>
#include <numeric>
#include <bitset>
#include <unordered_map>
#include <unordered_set>
#include <iomanip>
#include <queue>
#include <deque>
#include <stack>
#include <array>
#include <valarray>
#include <complex>
#include <random>
#include <functional>
#include <cassert>
#include <chrono>
#include <memory>

using namespace std;

#define int long long
//region TEMPLATE
////@formatter:off
typedef long long ll;
typedef unsigned long long ull;
template<typename T>void __read(T&a){cin >> a;}
template<typename T,typename ... Args>void __read(T&a,Args&... args){cin >> a;__read(args...);}
#define make(t, ...) t __VA_ARGS__; __read(__VA_ARGS__)
#define makev(a, n) VI a(n); for(auto&v:(a)) cin >> v;
#define ALL(x) (x).begin(), (x).end()
#define RALL(x) (x).rbegin(), (x).rend()
typedef vector<int> VI;
typedef vector<vector<int>> VVI;
typedef pair<int, int> PII;
constexpr ll M7 = 1000000007;
constexpr ll M9 = 1000000009;
constexpr ll MFFT = 998244353;
constexpr double PI = 3.14159265358979323846;
template<class T> void outv(const T&a,char d=' ', bool nl=false){for(auto&x:a)cout<<x<<d;if(nl)cout<<'\n';}
static mt19937_64 rnd(static_cast<unsigned>(chrono::steady_clock::now().time_since_epoch().count())); // NOLINT(cert-err58-cpp)
auto __fast_io__=(ios_base::sync_with_stdio(false),cin.tie(nullptr)); // NOLINT(cert-err58-cpp,cppcoreguidelines-interfaces-global-init)
////@formatter:on
//endregion


template<class T>
struct sparse {
    vector<vector<T>> table;
    vector<int32_t> lg;
    int32_t sz;

    inline T up(const T &a, const T &b) {
        return min(a, b);
//        return max(a, b);
    }

    template<class TIter>
    sparse(TIter begin, TIter end) {
        sz = distance(begin, end);
        lg.resize(sz);
        lg[0] = lg[1] = 0;
        for (int32_t i = 2; i < sz; ++i) {
            lg[i] = lg[i >> 1] + 1;
        }
        table.resize(lg[sz - 1] + 1);
        table[0].assign(begin, end);
        for (int32_t i = 1; i < table.size(); ++i) {
            table[i].resize(sz - (1 << i) + 1);
            int32_t sh = 1 << (i - 1);
            for (int32_t j = 0; j < table[i].size(); ++j) {
                table[i][j] = up(table[i - 1][j], table[i - 1][j + sh]);
            }
        }
    }

    T get(int32_t l, int32_t r) {
        int32_t i = lg[r - l];
        return up(table[i][l], table[i][r - (1 << i) + 1]);
    }
};

void solve(int32_t testNumber) {
    make(int, n);
    makev(p, n);
    sparse<int> st(ALL(p));
    VI pr(ALL(p));
    for (int i = 1; i < n; ++i) {
        pr[i] = max(pr[i], pr[i - 1]);
    }
    VI sf(ALL(p));
    for (int i = n - 2; i >= 0; --i) {
        sf[i] = max(sf[i], sf[i + 1]);
    }
    int j = n - 1;
    int nxJ = -2;
    for (int i = 0; i < j; ++i) {
        while (j > i + 2 && sf[j] < pr[i]) {
            j--;
            nxJ = -2;
        }
        if (sf[j] == pr[i] && j >= i + 2) {
            if (nxJ == -2) {
                nxJ = j;
                while (nxJ > 0 && sf[nxJ] == sf[j]) {
                    nxJ--;
                }
            }
            int l = max(i + 1, nxJ);
            int r = j - 1;
            while (l < r) {
                int mid = l + (r - l) / 2;
                if (st.get(i + 1, mid) > pr[i]) {
                    l = mid + 1;
                } else {
                    r = mid;
                }
            }
            if (pr[i] == st.get(i + 1, r)) {
                cout << "YES\n";
                cout << i + 1 << ' ' << r - i << ' ' << n - r - 1 << '\n';
                return;
            }
        }
    }
    cout << "NO\n";
}


int32_t main() {
    make(int32_t, testCount);
    for (int32_t testNumber = 0; testNumber < testCount; ++testNumber) {
        solve(testNumber);
    }
    return 0;
}