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;
const int N = 1e5;
int a[N+5];
struct ft{
	int ft[2*N+10];
	void init(){
		memset(ft, 0, sizeof(ft));
	}
	void update(int i, int v){
		for(;i<=2*N+5;i+=(i&-i)){
			ft[i] += v;
		}
	}
	int query(int i){
		int res = 0;
		for(;i>0;i-=(i&-i)) res+=ft[i];
		return res;
	}
}ft;
int main(){
	int n;
	scanf("%d", &n);
	for(int i=1;i<=n;i++){
		scanf("%d", &a[i]);
	}
	LL tot = (LL)n*(n+1)/2;
	LL m = tot/2+1;

	int lo = 1, hi = 1e9, ans = 0;
	while(lo <= hi){
		int mid = (lo+hi)/2;

		int cnt = 100001;
		LL res = 0;
		ft.init();
		for(int i=1;i<=n;i++){
			ft.update(cnt, 1);
			if(a[i] <= mid) cnt++;
			else cnt--;
			res += ft.query(cnt-1);
		}

		if(res >= m) ans = mid, hi = mid-1;
		else lo = mid+1;
	}
	printf("%d\n", ans);
}