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 maxn = 1e5 + 10;

ll n;
int arr[maxn];
int tree[maxn * 2];

int get_idx(int x) {
    return x + maxn;
}

void update(int x) {
    while(x < maxn * 2) {
        ++tree[x];
        x += x & -x;
    }
}

int query(int x) {
    int ret = 0;
    while(x) {
        ret += tree[x];
        x -= x & -x;
    }
    return ret;
}

int main(void) {
    scanf("%lld", &n);
    for(int i = 1; i <= n; ++i) scanf("%d", arr + i);
    int st = 1, ed = 1e9;
    int ans = INT_MAX;
    while(st <= ed) {
        int mid = st + ed >> 1;

        ll cnt = 0;
        memset(tree, 0, sizeof(tree));
        int base = 0;
        for(int i = 1; i <= n; ++i) {
            int now = arr[i] < mid ? -1 : 1;
            base += now;
            update(get_idx(now - base));
            cnt += query(get_idx(-base - 1));
        }

        //printf("%d %lld\n", mid, cnt);

        if(cnt <= n * (n + 1) / 4) { 
            st = mid + 1;
            ans = mid;              

        }
        else {
            ed = mid - 1;
        }
    }
    printf("%d", ans);
    return 0;
}