#include <iostream>
#include <vector>

using namespace std;
typedef long long ll;

template<typename T>
struct BinaryIndexedTree {
    vector<T> data;

    BinaryIndexedTree(int sz) {
        data.assign(++sz, 0);
    }

    T sum(int k) {
        T ret = 0;
        for (++k; k > 0; k -= k & -k) ret += data[k];
        return (ret);
    }

    void add(int k, T x) {
        for (++k; k < data.size(); k += k & -k) data[k] += x;
    }
};

ll solve(vector<int> v, int x) {
    ll ret = 0;

    int now = 150000;
    BinaryIndexedTree<int> BIT(300000);
    BIT.add(now, 1);
    for (auto a:v) {
        now += a >= x ? 1 : -1;
        ret += BIT.sum(now);
        BIT.add(now, 1);
    }

    return ret;
}

int main() {
    ll n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; i++)cin >> a[i];

    int ng = 1145141919, ok = 0;
    while (ng - ok > 1) {
        int mid = ok + (ng - ok) / 2;
        if (solve(a, mid) * 2 >= n * (n + 1) / 2)ok = mid;
        else ng = mid;
    }

    cout << ok << endl;
    return 0;
}
### Prompt

In Cpp, your task is to solve the following problem:

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>

using namespace std;
typedef long long ll;

template<typename T>
struct BinaryIndexedTree {
    vector<T> data;

    BinaryIndexedTree(int sz) {
        data.assign(++sz, 0);
    }

    T sum(int k) {
        T ret = 0;
        for (++k; k > 0; k -= k & -k) ret += data[k];
        return (ret);
    }

    void add(int k, T x) {
        for (++k; k < data.size(); k += k & -k) data[k] += x;
    }
};

ll solve(vector<int> v, int x) {
    ll ret = 0;

    int now = 150000;
    BinaryIndexedTree<int> BIT(300000);
    BIT.add(now, 1);
    for (auto a:v) {
        now += a >= x ? 1 : -1;
        ret += BIT.sum(now);
        BIT.add(now, 1);
    }

    return ret;
}

int main() {
    ll n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; i++)cin >> a[i];

    int ng = 1145141919, ok = 0;
    while (ng - ok > 1) {
        int mid = ok + (ng - ok) / 2;
        if (solve(a, mid) * 2 >= n * (n + 1) / 2)ok = mid;
        else ng = mid;
    }

    cout << ok << endl;
    return 0;
}
```