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;

const int N = 1e5 + 10;

int n, a[N], val[N], s[N];

int bit[N * 2 + 10];

void modify(int x) {
	x += n + 2;
	for( ; x <= 2 * n + 10; x += x & -x) ++ bit[x];
}

int query(int x) {
	x += n + 2;
	int res = 0;
	for( ; x ; x -= x & -x) res += bit[x];
	return res;
}

int check(int val) {
	// puts("go 1");
	for(int i = 0 ; i <= 2 * n + 10 ; ++ i) bit[i] = 0;
	for(int i = 1 ; i <= n ; ++ i) {
		if(a[i] <= val) s[i] = 1;
		else s[i] = -1;
		s[i] += s[i - 1];
	}
	long long res = 0, len = 1ll * n * (n + 1) / 2;
	modify(0);
	for(int i = 1 ; i <= n ; ++ i) {
		res += query(s[i] - 1);
		modify(s[i]);
	}
	return res >= len / 2 + 1;
}

int main() {
	scanf("%d", &n);
	for(int i = 1 ; i <= n ; ++ i) {
		scanf("%d", &a[i]);
		val[i] = a[i];
	}
	sort(val + 1, val + 1 + n);
	int k = unique(val + 1, val + 1 + n) - val - 1;
	int l = 1, r = k, ans = 0;
	while(l <= r) {
		int mid = (l + r) >> 1;
		if(check(val[mid])) {
			ans = val[mid];
			r = mid - 1;
		} else {
			l = mid + 1;
		}
	}
	printf("%d\n", ans);
}
