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;
typedef long long LL;
const int N = 100010;

LL tr[N << 1];
int a[N], n, s[N], tmp[N];
void update(int i, LL x) { for (; i <= n + N; i += i & -i) tr[i] += x; }
LL query(int i, LL ret = 0) { for (; i; i -= i & -i) ret += tr[i]; return ret; }

LL check(int num)
{
    LL ans = 0;
    memset(tr,0,sizeof tr);
    update(s[0]+N,1);
    for (int i = 1; i <= n; i++)
    {
        if (a[i] < num) s[i] = s[i - 1] - 1;
        if (a[i] >= num) s[i] = s[i - 1] + 1;
        ans += query(s[i] + N);
        update(s[i] + N, 1);
    }
    return ans;
}

int main()
{
    LL ans;
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> a[i];
    memcpy(tmp, a, sizeof a);
    sort(tmp + 1, tmp + n + 1);
    int cnt = unique(tmp + 1, tmp + n + 1) - tmp - 1;
    LL k = ((LL)n * (n + 1 ) / 2)/2;
    for (int l = 1, r = cnt, mid = l + r >> 1; l <= r; mid = l + r >> 1)
    {
        if (check(tmp[mid]) >= k) ans = tmp[mid], l = mid + 1;
        else r = mid - 1;
    }
    cout << ans << endl;
    return 0;
}