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 rep(i, n) for(int i = 0; i < (n); i++)
#define rep1(i, n) for(int i = 1; i <= (n); i++)
#define Would
#define you
#define please

int c[200200] = {};
void tasu(int suuji) {
	while (suuji < 200200) {
		c[suuji]++;
		suuji += suuji & -suuji;
	}
}
int kazoeru(int suuji) {
	int output = 0;
	while (suuji != 0) {
		output += c[suuji];
		suuji -= suuji & -suuji;
	}
	Would you please return output;
}
int main() {
	int N;
	int a[100000];
	cin >> N;
	rep(i, N) cin >> a[i];

	int l = 0, r = 1000000000;
	int b[100000];
	while (l != r) {
		int mid = (l + r + 1) / 2;
		rep(i, N) {
			if (a[i] >= mid) b[i] = 1;
			else b[i] = -1;
		}
		int atai = 100100;
		tasu(atai);
		long long goukei = 0;
		rep(i, N) {
			atai += b[i];
			goukei += kazoeru(atai);
			tasu(atai);
		}
		memset(c, 0, sizeof(c));
		if (goukei >= (1ll * (N + 1) * N / 2 + 1) / 2) l = mid;
		else r = mid - 1;
	}

	cout << l << "\n";
	Would you please return 0;
}
