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 <bits/stdc++.h>
using namespace std;

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

    sort(b.begin(), b.end());
    b.erase(unique(b.begin(), b.end()), b.end());

    auto f = [&](int k) {
        vector<int> s(2 * n + 1, 0);
        int sum = n; s[sum] = 1;
        long long ans = 0, add = 0;
        for (int i = 0; i < n; i++) {
            if (a[i] >= k) add -= s[--sum];
            else add += s[sum++];
            ans += add;
            s[sum]++;
        }
        return ans;
    };

    int ok = 0, ng = int(b.size());
    const long long x = ((1LL * n * (n + 1)) / 2);
    while (abs(ok - ng) > 1) {
        int mid = (ok + ng) / 2;
        if (f(b[mid]) * 2LL <= x) ok = mid;
        else ng = mid;
    }

    cout << b[ok] << '\n';

    return 0;
}