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;
#define ll long long

int a[100005], b[100005], psa[100005], bit[200005];
int n;

void update(int i) {
	for (; i < 200005; i += (i & -i))
		bit[i]++;
}

int query(int i) {
	int ans = 0;
	for (; i > 0; i -= (i & -i))
		ans += bit[i];
	return ans;
}

bool test(int m) {
	memset(bit, 0, sizeof bit);
	update(n + 1);
	psa[0] = 0;
	ll cnt = 0;
	for (int i = 1; i <= n; i++) {
		if (a[i-1] >= m)
			b[i] = 1;
		else
			b[i] = -1;
		psa[i] = psa[i - 1] + b[i];
		cnt += query(psa[i] + n + 1);
		update(psa[i] + n + 1);
	}
	return cnt >= (1LL * n * (n + 1) / 2 + 1) / 2;
}

int main() {
	scanf("%d", &n);
	int lo = 1, hi = 0;
	for (int i = 0; i < n; i++) {
		scanf("%d", &a[i]);
		hi = max(hi, a[i]);
	}
	while (lo < hi) {
		// printf("lo = %d, hi = %d\n", lo, hi);
		int mid = (lo + hi + 1) / 2;
		// printf("test(%d) = %d\n", mid, test(mid));
		if (test(mid))
			lo = mid;
		else
			hi = mid - 1;
	}
	printf("%d", lo);
}