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<stdio.h>
#include<ctype.h>
#include<string.h>
#include<algorithm>
#define il inline
#define rg register
#define lowbit(x) (x&(-x))
using namespace std;
typedef long long ll;
const int O=100010;
il int gi(){
	rg int o=0,fl=1;char ch=getchar();
	while(!isdigit(ch)&&ch!='-')ch=getchar();
	if(ch=='-')fl=-1,ch=getchar();
	while(isdigit(ch))o=o*10+ch-'0',ch=getchar();
	return o*fl;
}
ll n,a[O],tree[O<<1],sum[O],l=0x3f3f3f3f,r,ans;
il void add(int x,ll v){
	for(;x<=2e5;x+=lowbit(x))tree[x]+=v;
}
il int query(int x){
	int tot=0;
	for(;x;x-=lowbit(x))tot+=tree[x];
	return tot;
}
il bool check(int x){
	memset(tree,0,sizeof(tree));
	ans=0;
	for(int i=1;i<=n;++i)
		sum[i]=sum[i-1]+(a[i]>=x?1:-1);
	for(int i=0;i<=n;++i){
		ans+=query(sum[i]+O);
		add(sum[i]+O,1ll);
	}
	return ans>=n*(n+1)>>2;
}
int main( ){
	n=gi();
	for(int i=1;i<=n;++i){
		a[i]=gi();
		l=min(l,a[i]);
		r=max(r,a[i]);
	}
#define mid (l+r>>1)
	while(l<=r)
		if(check(mid))l=mid+1;
		else r=mid-1;
	printf("%lld\n",r);
#undef mid
	return 0;
}
