    #include<bits/stdc++.h>
    using namespace std;
    const int N=300010;
    int n;
    int a[N],h[N],cnt,b[N];
    int s[N<<2];
    int lowbit(int k){return k&-k;}
    void add(int k)
    {
    	for(int i=k;i<=600010;i+=lowbit(i))
    		s[i]++;
    }
    int query(int k)
    {
    	int res=0;
    	for(int i=k;i;i-=lowbit(i))
    		res+=s[i];
    	return res;
    }
    bool check(int mid)
    {
    	long long res=0;
    	for(int i=1;i<=n;i++)
    		b[i]=(a[i]<=mid)?1:-1;
    	int sum=300010;
    	memset(s,0,sizeof(s));
    	add(300010);
    	for(int i=1;i<=n;i++)
    	{
    		sum+=b[i];
    		add(sum); res+=query(sum-1);
    	}
    	if(res<((1ll*n*(n-1)/2+n)/2+1))
    		return 0;
    	else
    		return 1;
    }
    int main()
    {
    	scanf("%d",&n);
    	for(int i=1;i<=n;i++)
    		scanf("%d",&a[i]),h[i]=a[i];
    	sort(h+1,h+n+1); cnt=unique(h+1,h+n+1)-h-1;
    	for(int i=1;i<=n;i++)
    		a[i]=lower_bound(h+1,h+cnt+1,a[i])-h;
    	int l=1,r=cnt;
    	while(l<r)
    	{
    		int mid=l+r>>1;
    		if(check(mid))r=mid;
    		else
    			l=mid+1;
    	}
    	printf("%d",h[l]);
    	return 0;
    }
### Prompt

Generate 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<bits/stdc++.h>
    using namespace std;
    const int N=300010;
    int n;
    int a[N],h[N],cnt,b[N];
    int s[N<<2];
    int lowbit(int k){return k&-k;}
    void add(int k)
    {
    	for(int i=k;i<=600010;i+=lowbit(i))
    		s[i]++;
    }
    int query(int k)
    {
    	int res=0;
    	for(int i=k;i;i-=lowbit(i))
    		res+=s[i];
    	return res;
    }
    bool check(int mid)
    {
    	long long res=0;
    	for(int i=1;i<=n;i++)
    		b[i]=(a[i]<=mid)?1:-1;
    	int sum=300010;
    	memset(s,0,sizeof(s));
    	add(300010);
    	for(int i=1;i<=n;i++)
    	{
    		sum+=b[i];
    		add(sum); res+=query(sum-1);
    	}
    	if(res<((1ll*n*(n-1)/2+n)/2+1))
    		return 0;
    	else
    		return 1;
    }
    int main()
    {
    	scanf("%d",&n);
    	for(int i=1;i<=n;i++)
    		scanf("%d",&a[i]),h[i]=a[i];
    	sort(h+1,h+n+1); cnt=unique(h+1,h+n+1)-h-1;
    	for(int i=1;i<=n;i++)
    		a[i]=lower_bound(h+1,h+cnt+1,a[i])-h;
    	int l=1,r=cnt;
    	while(l<r)
    	{
    		int mid=l+r>>1;
    		if(check(mid))r=mid;
    		else
    			l=mid+1;
    	}
    	printf("%d",h[l]);
    	return 0;
    }
```