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
#include <bits/stdc++.h>

#define task "F"
#define all(v) (v).begin(), (v).end()
#define rep(i, l, r) for (int i = (l); i <= (r); ++i)
#define Rep(i, r, l) for (int i = (r); i >= (l); --i)
#define DB(X) { cerr << #X << " = " << (X) << '\n'; }
#define DB1(A, _) { cerr << #A << "[" << _ << "] = " << (A[_]) << '\n'; }
#define DB2(A, _, __) { cerr << #A << "[" << _ << "][" << __ << "] = " << (A[_][__]) << '\n'; }
#define DB3(A, _, __, ___) { cerr << #A << "[" << _ << "][" << __ << "][" << ___ << "] = " << (A[_][__][___]) << '\n'; }
#define PR(A, l, r) { cerr << '\n'; rep(_, l, r) DB1(A, _); cerr << '\n';}
#define SZ(x) ((int)(x).size())
#define pb push_back
#define eb emplace_back
#define pf push_front
#define F first
#define S second
#define by(x) [](const auto& a, const auto& b) { return a.x < b.x; } // sort(arr, arr + N, by(a));
#define next ___next
#define prev ___prev
#define y1 ___y1
#define left ___left
#define right ___right
#define y0 ___y0
#define div ___div
#define j0 ___j0
#define jn ___jn

using ll = long long;
using ld = long double;
using ull = unsigned long long;
using namespace std;
typedef pair<int, int> ii;
typedef pair<ii, int> iii;
typedef vector<int> vi;
typedef vector<ii> vii;
typedef vector<ll> vl;
const int N = 2e5 + 2;
int T, n, a[N], mn[N][18], pre[N], suf[N];
int query(int l, int r)
{
    int k = 31 - __builtin_clz(r - l + 1);
    return min(mn[l][k], mn[r - (1 << k) + 1][k]);
}
vi pos[N];
int main()
{
#ifdef HynDuf
    freopen(task".in", "r", stdin);
    //freopen(task".out", "w", stdout);
#else
    ios_base::sync_with_stdio(false); cin.tie(nullptr);
#endif
    cin >> T;
    while (T--)
    {
        cin >> n;
        vi com;
        rep(i, 1, n) cin >> a[i], com.eb(a[i]);
        sort(all(com));
        com.erase(unique(all(com)), com.end());
        rep(i, 1, n) a[i] = lower_bound(all(com), a[i]) - com.begin() + 1;
        rep(i, 1, SZ(com)) pos[i].clear();
        rep(i, 1, n) pos[a[i]].eb(i);
        rep(i, 1, n) mn[i][0] = a[i];
        suf[n + 1] = 0;
        rep(i, 1, n) pre[i] = max(pre[i - 1], a[i]);
        Rep(i, n, 1) suf[i] = max(suf[i + 1], a[i]);
        rep(i, 1, 17) for (int j = 1; j + (1 << i) - 1 <= n; j++) mn[j][i] = min(mn[j][i - 1], mn[j + (1 << (i - 1))][i - 1]);
        rep(i, 1, SZ(com)) if (SZ(pos[i]) >= 3)
        {
            rep(j, 1, SZ(pos[i]) - 2)
            {
                int p = pos[i][j];
                int l = pos[i][0] + 1, r = p - 1;
                while (l <= r)
                {
                    int m = (l + r) >> 1;
                    if (query(m, p) >= i) r = m - 1;
                    else l = m + 1;
                }
                int l1 = l;
                l = p + 1, r = pos[i].back() - 1;
                while (l <= r)
                {
                    int m = (l + r) >> 1;
                    if (query(p, m) >= i) l = m + 1;
                    else r = m - 1;
                }
                int r1 = r;
                if (pre[l1 - 1] == i && suf[r1 + 1] == i && query(l1, r1) == i)
                {
                    cout << "YES\n" << l1 - 1 << ' ' << r1 - l1 + 1 << ' ' << n - r1 << '\n';
                    goto Next;
                }
            }
        }
        cout << "NO\n";
        Next:;
    }
    return 0;
}
