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 <iomanip>
#include <algorithm>
#include <map>
#include <vector>
#include <string>

using namespace std;
typedef long long ll;
const int MAXN = 100100;

int N;
int arr[MAXN];
int p[MAXN];
int csum[2*MAXN];

bool good (int cc)
{
    for (int i = 0; i < N; i++)
    {
        if (arr[i] >= cc)
            p[i] = 1;
        else
            p[i] = -1;
    }
    for (int i = 0; i < 2 * MAXN; i++)
        csum[i] = 0;

    int cs = MAXN;
    csum[cs] = 1;
    int ctot = 1;
    ll res = 0;
    for (int i = 0; i < N; i++)
    {
        if (p[i] == -1)
        {
            ctot -= csum[cs];
            cs--;
        }
        else
        {
            ctot += csum[cs+1];
            cs++;
        }
        res += ctot;
        csum[cs]++;
        ctot++;
    }
    //cout << cc << " " << res << "\n";
    return res * 2 >= (N * ((ll) N + 1)) / 2;
}

int main()
{
    ios_base::sync_with_stdio(0);
    cin >> N;
    for (int i = 0; i < N; i++)
        cin >> arr[i];

    int lo = 0, hi = 1e9;
    while (lo < hi)
    {
        int mid = (lo + hi + 1) / 2;
        if (good (mid))
            lo = mid;
        else
            hi = mid - 1;
    }
    cout << lo << "\n";
}