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>
#define LL long long
using namespace std;
int n;
int arr[100100];
inline int lowbit(int x)
{
    return x & -x;
}
int sum[200100];
void add1(int x)
{
    while(x <= 2 * n + 1)
    {
        sum[x] ++;
        x += lowbit(x);
    }
}
int query(int x)
{
    int ret = 0;
    while(x)
    {
        ret += sum[x];
        x -= lowbit(x);
    }
    return ret;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n;
    for(int i = 1; i <= n; i ++)
        cin >> arr[i];
    int l = 0, r = 1e9;
    LL wtd = ((LL)n * n + n) / 2 + 1 >> 1;
    while(l != r)
    {
        int mid = (l + r) / 2 + 1;
        LL tot = 0;
        memset(sum, 0, sizeof(sum));
        int tmp = n + 1;
        add1(tmp);
        for(int i = 1; i <= n; i ++)
        {
            tmp += (arr[i] >= mid ? 1 : -1);
            tot += query(tmp);
            add1(tmp);
        }
//        if(mid == 8)
//            cout << tot << endl;
        if(tot >= wtd)
            l = mid;
        else
            r = mid - 1;
    }
    cout << l << endl;
    return 0;
}
