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++)
int a[100005],l=0,r=1e9+7;
long long n,ac;
struct BIT{
	int d[(1<<18)+1] = {};
	void ope(int x){
		for(int y=x;y;y-=y&-y) ac -= d[y];
		for(;x<=1<<18;x+=x&-x) d[x]++;
	}
};
int main(){
	scanf("%d", &n);
	rep(i,n) scanf("%d", a+i);
	while(l+1<r){
		int m=(l+r)/2, s=n+1;
		BIT bt;bt.ope(s);
		rep(i,n) bt.ope(s+=a[i]<m?-1:1);
		(ac+n*(n+1)/2<=n*(n+1)/4?l:r)=m;ac=0;
	}
	printf("%d\n", l);
}