#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mn=1e5+10;
int a[mn],b[mn],c[mn*2];
int n;
bool leq(int x){
    int i;
    for(i=1;i<=n;i++)b[i]=a[i]>x?1:-1;
    for(i=1;i<=n;i++)b[i]+=b[i-1];
    memset(c,0,sizeof(c));
    for(i=1;i<=n;i++)c[mn+b[i]]++;
    ll tot=0;
    int cur=0;
    for(i=0;i<mn;i++)cur+=c[i];
    for(i=0;i<n;i++){
        tot+=cur;
        if(b[i+1]<b[i]){
            cur--;
            c[b[i+1]+mn]--;
            cur-=c[b[i]+mn-1];
        }
        else{
            c[b[i+1]+mn]--;
            cur+=c[b[i+1]+mn-1];
        }
    }
    //tot+=cur;
    return (tot>=(1LL*n*(n+1)/2)/2+1);
}
int main(){
    int i;
    scanf("%d",&n);
    for(i=1;i<=n;i++)scanf("%d",a+i);
    int l=1,r=1e9;
    while(l!=r){
        int mid=l+r>>1;
        if(!leq(mid))l=mid+1;
        else r=mid;
    }
    printf("%d",l);
}
### 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;
typedef long long ll;
const int mn=1e5+10;
int a[mn],b[mn],c[mn*2];
int n;
bool leq(int x){
    int i;
    for(i=1;i<=n;i++)b[i]=a[i]>x?1:-1;
    for(i=1;i<=n;i++)b[i]+=b[i-1];
    memset(c,0,sizeof(c));
    for(i=1;i<=n;i++)c[mn+b[i]]++;
    ll tot=0;
    int cur=0;
    for(i=0;i<mn;i++)cur+=c[i];
    for(i=0;i<n;i++){
        tot+=cur;
        if(b[i+1]<b[i]){
            cur--;
            c[b[i+1]+mn]--;
            cur-=c[b[i]+mn-1];
        }
        else{
            c[b[i+1]+mn]--;
            cur+=c[b[i+1]+mn-1];
        }
    }
    //tot+=cur;
    return (tot>=(1LL*n*(n+1)/2)/2+1);
}
int main(){
    int i;
    scanf("%d",&n);
    for(i=1;i<=n;i++)scanf("%d",a+i);
    int l=1,r=1e9;
    while(l!=r){
        int mid=l+r>>1;
        if(!leq(mid))l=mid+1;
        else r=mid;
    }
    printf("%d",l);
}
```