#include<bits/stdc++.h>
using namespace std;
#define RI register int
int read() {
	int q=0;char ch=' ';
	while(ch<'0'||ch>'9') ch=getchar();
	while(ch>='0'&&ch<='9') q=q*10+ch-'0',ch=getchar();
	return q;
}
typedef long long LL;
const int N=100005,inf=0x3f3f3f3f;
int n,js,a[N],s[N],b[N],tr[N];
int lowbit(int x) {return x&(-x);}
void add(int x) {while(x<=js) ++tr[x],x+=lowbit(x);}
int query(int x) {
	int re=0;
	while(x) re+=tr[x],x-=lowbit(x);
	return re;
}
int check(int x) {
	for(RI i=1;i<=n;++i)
		s[i]=s[i-1]+(a[i]>=x?1:-1),b[i]=s[i];
	b[n+1]=0,sort(b+1,b+2+n);
	js=1;for(RI i=2;i<=n+1;++i) if(b[i]!=b[js]) b[++js]=b[i];
	LL res=0,num=1LL*n*(n-1)/2+n;
	for(RI i=1;i<=js;++i) tr[i]=0;
	for(RI i=0;i<=n;++i) {
		int k=lower_bound(b+1,b+1+js,s[i])-b;
		res+=query(k),add(k);
	}
	return res>=(num+1)/2;
}
int main()
{
	int l=inf,r=0,ans;
	n=read();
	for(RI i=1;i<=n;++i)
		a[i]=read(),l=min(a[i],l),r=max(a[i],r);
	while(l<=r) {
		int mid=(l+r)>>1;
		if(check(mid)) ans=mid,l=mid+1;
		else r=mid-1;
	}
	printf("%d\n",ans);
	return 0;
}
### 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<bits/stdc++.h>
using namespace std;
#define RI register int
int read() {
	int q=0;char ch=' ';
	while(ch<'0'||ch>'9') ch=getchar();
	while(ch>='0'&&ch<='9') q=q*10+ch-'0',ch=getchar();
	return q;
}
typedef long long LL;
const int N=100005,inf=0x3f3f3f3f;
int n,js,a[N],s[N],b[N],tr[N];
int lowbit(int x) {return x&(-x);}
void add(int x) {while(x<=js) ++tr[x],x+=lowbit(x);}
int query(int x) {
	int re=0;
	while(x) re+=tr[x],x-=lowbit(x);
	return re;
}
int check(int x) {
	for(RI i=1;i<=n;++i)
		s[i]=s[i-1]+(a[i]>=x?1:-1),b[i]=s[i];
	b[n+1]=0,sort(b+1,b+2+n);
	js=1;for(RI i=2;i<=n+1;++i) if(b[i]!=b[js]) b[++js]=b[i];
	LL res=0,num=1LL*n*(n-1)/2+n;
	for(RI i=1;i<=js;++i) tr[i]=0;
	for(RI i=0;i<=n;++i) {
		int k=lower_bound(b+1,b+1+js,s[i])-b;
		res+=query(k),add(k);
	}
	return res>=(num+1)/2;
}
int main()
{
	int l=inf,r=0,ans;
	n=read();
	for(RI i=1;i<=n;++i)
		a[i]=read(),l=min(a[i],l),r=max(a[i],r);
	while(l<=r) {
		int mid=(l+r)>>1;
		if(check(mid)) ans=mid,l=mid+1;
		else r=mid-1;
	}
	printf("%d\n",ans);
	return 0;
}
```