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 <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define ll long long
#define lowbit(a) a&(-a)
#define INF 0x3f3f3f3f
using namespace std;

inline int read(){
	int sum=0,f=1;char ch=getchar();
	while(ch>'9' || ch<'0'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0' && ch<='9'){sum=sum*10+ch-'0';ch=getchar();}
	return f*sum;
}

const int N=100001;
ll n,a[N];ll s[N];
ll c[N],tot;ll t[(N<<1)+5];

inline ll ask(int x){
	ll ret=0;
	for(int i=x;i;i-=lowbit(i)) ret+=t[i];
	return ret;
}

inline void add(int x,int k){
	for(int i=x;i<=(N<<1);i+=lowbit(i)) t[i]+=k;
}

inline bool check(int mid){
	memset(t,0,sizeof(t));s[0]=0;ll sum=0;
	for(int i=1;i<=n;i++) s[i]=s[i-1]+(a[i]>=mid? 1:-1);
	for(int i=0;i<=n;i++){
		sum+=ask(s[i]+N);add(s[i]+N,1);
	}
	return sum>=n*(n+1)/4;
}

int main(){
	n=read();
	for(int i=1;i<=n;i++) a[i]=read(),c[i]=a[i];
	sort(c+1,c+n+1);int m=unique(c+1,c+n+1)-c-1;
	for(int i=1;i<=n;i++) a[i]=lower_bound(c+1,c+m+1,a[i])-c;
	int l=1,r=n;
	while(l<=r){
		int mid=(l+r)>>1;
		if(check(mid)) l=mid+1;
		else r=mid-1;
	}
	printf("%lld\n",c[r]);
	return 0;
}
