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 int long long

using namespace std;

int a[100010],n;

int solve(int m) 
{
    vector<int> s(2 * n + 1);
    
    int sum = n;
    int result = 0;
    s[sum] = 1;
    int add = 1;

    for(int i=1;i<=n;i++)
    {
        if (a[i] < m)
            add -= s[sum], sum--;
        else
            sum++, add += s[sum];
            
        result += add;
        s[sum]++;
        add++;
    }
    return result;
}

int32_t main()
{
    cin >> n;

    for(int i=1;i<=n;i++)
        cin >> a[i];

    int total = n*(n+1)/2;
    total = total/2 + 1;
    total = n*(n+1)/2-total;

    int l = *min_element(a+1,a+n+1);
    int h = *max_element(a+1,a+n+1);

    if(l == h)
    {
        cout << l << endl;
        return 0;
    }

    //cout << total << endl;

    int ans = INT_MIN;

    while(l <= h)
    {
        int mid = (l+h)/2;

        int temp = solve(mid);

        //cout << mid << " " << temp << endl;

        if(temp < total)
            h = mid-1;
        else
        {
            l = mid+1;

            ans = max(ans,mid);
        }
    }

    cout << ans << endl;
}