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

struct BinaryIndexedTree {
    vector<int> data;
    int n;
    BinaryIndexedTree(int n) : data(n+1), n(n+1) {}

    void update(int i,long x){
        for(; i <= n; i += i & -i) data[i] += x;
    }

    long query(int i){
        long ret = 0;
        for(; i > 0; i -= i & -i) ret += data[i];
        return ret;
    }
};

int main() {
    cin.tie(0); ios_base::sync_with_stdio(false);
    long n; cin >> n;
    long a[100010]; for(int i = 0; i < n; ++i) cin >> a[i];

    int l = 0, r = 2e9;
    while(r - l > 1) {
        long mid = (l + r) / 2;
        BinaryIndexedTree bit(n * 2);
        long sum = n + 1, num = 0;
        bit.update(sum, 1);

        for(int i = 0; i < n; ++i) {
            sum += (a[i] <= mid ? 1 : -1);
            num += bit.query(sum - 1);
            bit.update(sum, 1);
        }

        if(num > n * (n + 1) / 2 - num) r = mid;
        else l = mid;
    }

    cout << r << endl;
    return 0;
}