#include <stdio.h>
#define Maxn 100005
int n,ans,a[Maxn],rmq[Maxn*4],s[Maxn];
void add(int x) {for (;x<Maxn*4;x+=x&-x) rmq[x]++;}
int sum(int x) {int res=0;for (;x;x-=x&-x) res+=rmq[x];return res;}
int main()
{
	scanf("%d",&n);
	for (int i=0;i<n;++i) scanf("%d",a+i);
	for (int l=0,r=1e9;l<=r;)
	{
		__builtin_memset(rmq,0,sizeof(rmq));
		int mid=l+r>>1;
		long long S=0;
		add(n+1);
		for (int i=0;i<n;++i)
		{
			s[i]=(i?s[i-1]:0)+(a[i]>=mid?1:-1);
			S+=sum(s[i]+n+1);
			add(s[i]+n+1);
		}
		if (S>=1LL*n*(n+1)/2-S) ans=mid,l=mid+1;
		else r=mid-1;
	}
	printf("%d\n",ans);
	return 0;
}
### Prompt

Your task is to create a Cpp solution to the following problem:

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

### Response

```cpp
#include <stdio.h>
#define Maxn 100005
int n,ans,a[Maxn],rmq[Maxn*4],s[Maxn];
void add(int x) {for (;x<Maxn*4;x+=x&-x) rmq[x]++;}
int sum(int x) {int res=0;for (;x;x-=x&-x) res+=rmq[x];return res;}
int main()
{
	scanf("%d",&n);
	for (int i=0;i<n;++i) scanf("%d",a+i);
	for (int l=0,r=1e9;l<=r;)
	{
		__builtin_memset(rmq,0,sizeof(rmq));
		int mid=l+r>>1;
		long long S=0;
		add(n+1);
		for (int i=0;i<n;++i)
		{
			s[i]=(i?s[i-1]:0)+(a[i]>=mid?1:-1);
			S+=sum(s[i]+n+1);
			add(s[i]+n+1);
		}
		if (S>=1LL*n*(n+1)/2-S) ans=mid,l=mid+1;
		else r=mid-1;
	}
	printf("%d\n",ans);
	return 0;
}
```