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;

ll n;
vector<long long> bit(200010,0);
void add(int a, int w) {
	for (int x = a;x <= 2*n+1;x += x&-x) bit[x] += w;
}
int sum(int a) {
	int ret = 0;
	for (int x = a;x > 0;x -= x&-x) ret += bit[x];
	return ret;
}

int main() {
    vector<int> a(100010,0);
    cin >> n;
    for (int i = 0;i < n;++i) cin >> a[i];

    int min = 0,max = 1000000010;
    ll m = n*(n+1)/2;
    m = (m+1)/2;
    while (1) {
        ll ret = 0;
        int mid = min+(max-min)/2;
        vector<int> b(n+1);
        for (int i = 1;i <= n;++i) {
            b[i] = ((a[i-1] < mid) ? -1 : 1);
        }
        for (int i = 1;i <= n;++i) b[i] += b[i-1];
        bit = vector<ll>(200010,0);
        for (int i = n;i >= 0;--i) {
            ret += n-i-sum(n+b[i]);
            add(n+1+b[i],1);
        }
        if (ret < m) max = mid;
        else if (min == mid) break;
        else min = mid;
    }
    cout << min << endl;
}