#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<set>
#define MN 100005
using namespace std;
int n,tt,a[MN],num[MN],tmp[MN],tmpx[MN];
long long tx;
void mergesort(int ll,int rr){
	if(ll==rr)return;
	int mid=(ll+rr)>>1;
	mergesort(ll,mid),mergesort(mid+1,rr);
	int tl=ll,tr=mid+1,now=0;
	while(tl<=mid&&tr<=rr){
		if(tmp[tl]<tmp[tr])tmpx[++now]=tmp[tl++];
		else {
			tx+=tl-ll;
			tmpx[++now]=tmp[tr++];
		}
	}
	while(tl<=mid)tmpx[++now]=tmp[tl++];
	while(tr<=rr)tx+=tl-ll,tmpx[++now]=tmp[tr++];
	for(int i=ll,j=1;i<=rr;i++,j++)tmp[i]=tmpx[j];
}
bool check(int mid){
	for(int i=1;i<=n;i++)
		tmp[i]=tmp[i-1]+(a[i]<=mid?1:-1);
	tx=0;mergesort(0,n);
	return tx>=(1ll*n*(n+1)/4+1);
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d",&a[i]),num[++tt]=a[i];sort(num+1,num+tt+1);tt=unique(num+1,num+n+1)-num;
	int l=1,r=tt-1,ans=tt-1;
	while(l<=r){
		int mid=(l+r)>>1;
		if(check(num[mid]))ans=mid,r=mid-1;
		else l=mid+1;
	}printf("%d\n",num[ans]);
}
### Prompt

Create a solution in Cpp for 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<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<set>
#define MN 100005
using namespace std;
int n,tt,a[MN],num[MN],tmp[MN],tmpx[MN];
long long tx;
void mergesort(int ll,int rr){
	if(ll==rr)return;
	int mid=(ll+rr)>>1;
	mergesort(ll,mid),mergesort(mid+1,rr);
	int tl=ll,tr=mid+1,now=0;
	while(tl<=mid&&tr<=rr){
		if(tmp[tl]<tmp[tr])tmpx[++now]=tmp[tl++];
		else {
			tx+=tl-ll;
			tmpx[++now]=tmp[tr++];
		}
	}
	while(tl<=mid)tmpx[++now]=tmp[tl++];
	while(tr<=rr)tx+=tl-ll,tmpx[++now]=tmp[tr++];
	for(int i=ll,j=1;i<=rr;i++,j++)tmp[i]=tmpx[j];
}
bool check(int mid){
	for(int i=1;i<=n;i++)
		tmp[i]=tmp[i-1]+(a[i]<=mid?1:-1);
	tx=0;mergesort(0,n);
	return tx>=(1ll*n*(n+1)/4+1);
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d",&a[i]),num[++tt]=a[i];sort(num+1,num+tt+1);tt=unique(num+1,num+n+1)-num;
	int l=1,r=tt-1,ans=tt-1;
	while(l<=r){
		int mid=(l+r)>>1;
		if(check(num[mid]))ans=mid,r=mid-1;
		else l=mid+1;
	}printf("%d\n",num[ans]);
}
```