#include <bits/stdc++.h>
using namespace std;
int a[100100],b[100100],c[200100],n;
 
int lowbit(int x)
{
    return x&(-x);
}
 
void add(int x,int val)
{
    x=x+(n+1);
    int i;
    for (i=x;i<=2*n+1;i+=lowbit(i)) c[i]+=val;
}
 
long long sum(int x)
{
    x+=(n+1);
    int i;
    long long ans=0;
    for (i=x;i>0;i-=lowbit(i)) ans+=c[i];
    return ans;
}
 
long long work(int num)
{
    int i,tmp=0;long long ans=0;
    for (i=1;i<=n;i++)
    {
        if (a[i]<=num) tmp++; else tmp--;
        ans+=sum(tmp-1);
        add(tmp,1);
    }
    tmp=0;
    for (i=1;i<=n;i++)
    {
        if (a[i]<=num) tmp++; else tmp--;
        add(tmp,-1);
    }
    return ans;
}
 
int main()
{
    int i;
    scanf("%d",&n);
    for (i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        b[i]=a[i];
    }
    memset(c,0,sizeof(c));
    sort(b+1,b+1+n);
    int m=unique(b+1,b+1+n)-b-1;
    int l=1,r=m,ans;
    long long target=(long long)n*(n+1)/2;target=target/2+1;
    add(0,1);
    while (l<=r)
    {
        int mid=(l+r)/2;
        if (work(b[mid])>=target) {ans=b[mid];r=mid-1;}
        else l=mid+1;
    }
    printf("%d\n",ans);
    return 0;
}
### Prompt

Please create a solution in Cpp 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;
int a[100100],b[100100],c[200100],n;
 
int lowbit(int x)
{
    return x&(-x);
}
 
void add(int x,int val)
{
    x=x+(n+1);
    int i;
    for (i=x;i<=2*n+1;i+=lowbit(i)) c[i]+=val;
}
 
long long sum(int x)
{
    x+=(n+1);
    int i;
    long long ans=0;
    for (i=x;i>0;i-=lowbit(i)) ans+=c[i];
    return ans;
}
 
long long work(int num)
{
    int i,tmp=0;long long ans=0;
    for (i=1;i<=n;i++)
    {
        if (a[i]<=num) tmp++; else tmp--;
        ans+=sum(tmp-1);
        add(tmp,1);
    }
    tmp=0;
    for (i=1;i<=n;i++)
    {
        if (a[i]<=num) tmp++; else tmp--;
        add(tmp,-1);
    }
    return ans;
}
 
int main()
{
    int i;
    scanf("%d",&n);
    for (i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        b[i]=a[i];
    }
    memset(c,0,sizeof(c));
    sort(b+1,b+1+n);
    int m=unique(b+1,b+1+n)-b-1;
    int l=1,r=m,ans;
    long long target=(long long)n*(n+1)/2;target=target/2+1;
    add(0,1);
    while (l<=r)
    {
        int mid=(l+r)/2;
        if (work(b[mid])>=target) {ans=b[mid];r=mid-1;}
        else l=mid+1;
    }
    printf("%d\n",ans);
    return 0;
}
```