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;
typedef long long ll;

const int MAX_N = 1e5 + 10;

ll N;
ll A[MAX_N], sum_A[MAX_N], bit[2 * MAX_N];

ll sum(ll i) {
    ll s = 0;
    while (i > 0) {
        s += bit[i];
        i -= i & -i;
    }
    return s;
}

void add(ll i, ll x) {
    while (i <= 2 * N + 1) {
        bit[i] += x;
        i += i & -i;
    }
}

int main() {
    cin >> N;
    for (int i = 1; i <= N; i++) cin >> A[i];
    if (N == 1) {
        cout << A[1] << endl;
        return 0;
    }
    ll left = 1, right = 1e9 + 1;
    while (right - left > 1) {
        ll mid = (right + left) / 2;
        for (int i = 1; i <= N; i++) {
            sum_A[i] = A[i] < mid ? -1 : 1;
        }
        for (int i = 1; i <= N; i++) {
            sum_A[i] += sum_A[i - 1];
        }
        for (int i = 0; i <= 2 * N + 1; i++) bit[i] = 0;
        ll cnt = 0;
        for (int i = 0; i <= N; i++) {
            cnt += sum(sum_A[i] + N + 1);
            add(sum_A[i] + N + 1, 1);
        }
        if (cnt >= N * (N + 1) / 4) left = mid;
        else right = mid;
    }
    cout << left << endl;
    return 0;
}