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;
int n, a[100020];
int c[200020];
void R(int x, int y) {
	for (; x <= 2 * n + 10; x += x & -x) {
		c[x] += y;
	}
}
int G(int x) {
	int re = 0;
	for (; x > 0; x -= x & -x) {
		re += c[x];
	}
	return re;
}
bool ok(int M) {
	memset(c, 0, sizeof c);
	int s = n + 1;
	long long cnt = 0;
	for (int i = 0; i < n; i++) {
		R(s, 1);
		if (a[i] >= M) {
			s++;
		} else {
			s--;
		}
		cnt += G(s);
	}
	return cnt * 2 >= (long long)n * (n + 1) / 2;
}
int main() {
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		scanf("%d", &a[i]);
	}
	int L = 0, R = 1e9 + 1;
	while (L < R - 1) {
		int M = (L + R) / 2;
		if (ok(M)) {
			L = M;
		} else {
			R = M;
		}
	}
	printf("%d\n", L);
	return 0;
}