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<cstdio>
#include<algorithm>
using namespace std;const int N=1e5+10;typedef long long ll;
struct data
{
	int val;int pos;
	friend bool operator <(data a,data b)
		{return (a.val==b.val)?a.pos<b.pos:a.val<b.val;}
}a[N],b[N];int n;ll tot;int sum[N];
struct treearray
{
	int ta[2*N];
	inline void clear(){for(int i=1;i<=2*n+10;i++)ta[i]=0;}
	inline void c(int x){for(;x<=2*n+10;x+=x&(-x))ta[x]++;}
	inline int q(int x){int r=0;for(;x>0;x-=x&(-x))r+=ta[x];return r;}
}ta;
inline bool jud(int x)
{
	ll ret=0;ta.clear();
	for(int i=1;i<=n;i++)sum[i]=(a[x]<a[i])?-1:1;
	//for(int i=1;i<=n;i++)printf("%d ",sum[i]);printf("\n");
	for(int i=1;i<=n;i++)sum[i]+=sum[i-1];ta.c(n+1);
	for(int i=1;i<=n;i++){ret+=ta.q(sum[i]+n);ta.c(sum[i]+n+1);}
	//printf("jud(%d)=%lld\n",a[x].val,ret);
	return ret>=(tot/2)+1;
}
int main()
{
	scanf("%d",&n);tot=(ll)n*(n-1)/2+n;
	for(int i=1;i<=n;i++)scanf("%d",&a[i].val),a[i].pos=i;
	for(int i=1;i<=n;i++)b[i]=a[i];sort(b+1,b+n+1);int l=1;int r=n;
	while(l!=r){int mid=(l+r)/2;if(jud(b[mid].pos))r=mid;else l=mid+1;}
	printf("%d",b[l].val);return 0;
}