#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

struct BIT {
    int n;
    vector<int> v;

    BIT(int n) : v(n+1) {
        this->n = n;
    }

    void add(int a, int x) {
        for (int i = a; i <= n; i += i & -i)
            v[i] += x;
    }

    int sum(int a) {
        int res = 0;
        for (int i = a; i > 0; i -= i & -i)
            res += v[i];
        return res;
    }
};

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    int N; cin >> N;
    vector<int> a(N);
    int maxA = 0;
    for (int i = 0; i < N; i++) {
        cin >> a[i];
        maxA = max(maxA, a[i]);
    }
    long long ml = (1LL + N) * N / 2;

    int low = 0, high = maxA+1;
    while (high - low > 1) {
        int mid = (low + high) / 2;

        BIT bit(N*2+2);
        int cs = N+1;
        long long res = 0;
        bit.add(cs, 1);
        for (int i = 0; i < N; i++) {
            cs += a[i] < mid ? -1 : 1;
            res += bit.sum(cs);
            bit.add(cs, 1);
        }

        if (res >= (ml+1)/2)
            low = mid;
        else
            high = mid;
    }
    cout << low << "\n";
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

We will define the median of a sequence b of length M, as follows:

* Let b' be the sequence obtained by sorting b in non-decreasing order. Then, the value of the (M / 2 + 1)-th element of b' is the median of b. Here, / is integer division, rounding down.



For example, the median of (10, 30, 20) is 20; the median of (10, 30, 20, 40) is 30; the median of (10, 10, 10, 20, 30) is 10.

Snuke comes up with the following problem.

You are given a sequence a of length N. For each pair (l, r) (1 \leq l \leq r \leq N), let m_{l, r} be the median of the contiguous subsequence (a_l, a_{l + 1}, ..., a_r) of a. We will list m_{l, r} for all pairs (l, r) to create a new sequence m. Find the median of m.

Constraints

* 1 \leq N \leq 10^5
* a_i is an integer.
* 1 \leq a_i \leq 10^9

Input

Input is given from Standard Input in the following format:


N
a_1 a_2 ... a_N


Output

Print the median of m.

Examples

Input

3
10 30 20


Output

30


Input

1
10


Output

10


Input

10
5 9 5 9 8 9 3 5 4 3


Output

8

### Response

```cpp
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

struct BIT {
    int n;
    vector<int> v;

    BIT(int n) : v(n+1) {
        this->n = n;
    }

    void add(int a, int x) {
        for (int i = a; i <= n; i += i & -i)
            v[i] += x;
    }

    int sum(int a) {
        int res = 0;
        for (int i = a; i > 0; i -= i & -i)
            res += v[i];
        return res;
    }
};

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    int N; cin >> N;
    vector<int> a(N);
    int maxA = 0;
    for (int i = 0; i < N; i++) {
        cin >> a[i];
        maxA = max(maxA, a[i]);
    }
    long long ml = (1LL + N) * N / 2;

    int low = 0, high = maxA+1;
    while (high - low > 1) {
        int mid = (low + high) / 2;

        BIT bit(N*2+2);
        int cs = N+1;
        long long res = 0;
        bit.add(cs, 1);
        for (int i = 0; i < N; i++) {
            cs += a[i] < mid ? -1 : 1;
            res += bit.sum(cs);
            bit.add(cs, 1);
        }

        if (res >= (ml+1)/2)
            low = mid;
        else
            high = mid;
    }
    cout << low << "\n";
}
```