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>
#define lowbit(x) ((x)&(-x))
using namespace std;

const int maxn=1000003;

long long sum[maxn];
int n;
inline void update(int x){
	for(int i=x;i<maxn;i+=lowbit(i))
		++sum[i];
}
inline int query(int x){
	static int rt;rt=0;
	for(int i=x;i>0;i-=lowbit(i))
		rt+=sum[i];
	return rt;
}
long long KK=5e5,a[maxn],pre[maxn];
inline bool check(int x){
	static long long res;res=0;
	pre[0]=0;
	for(int i=1;i<=n;++i)
		pre[i]=pre[i-1]+(a[i]>=x?1:-1);
	memset(sum,0,sizeof(sum));
	for(int i=0;i<=n;++i){
		res+=query(pre[i]+KK);
		update(pre[i]+KK);
	}
	return res>=1ll*n*(n+1)/4;
}

int main(){
	scanf("%d",&n);
	if(n==1){
		scanf("%lld",&a[1]);
		printf("%lld\n",a[1]);
		return 0;
	}
	for(int i=1;i<=n;++i)
		scanf("%lld",&a[i]);
	int l=-1,r=1e9+7;
	while(l<r-1){
		int mid=(l+r)/2;
		if(check(mid)) l=mid;
		else r=mid;
	}
	printf("%d\n",l);
	return 0;
}
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