#include <bits/stdc++.h>
#define IDX 131072
using namespace std;
int ans;
int N;
int m[111111],t[111111];
int main(){
  cin>>N;
  for(int i=0;i<N;i++){
    cin>>m[i];
    t[i]=m[i];
  }
  sort(t,t+N);
  int a=0,b=N;
  while(a<b){
    int c=(a+b)/2;
    int d=t[c];
    vector<pair<int,int>>v(N+1);
    vector<int>dp(IDX*2);
    for(int i=1;i<=N;i++){
      if(m[i-1]<d)v[i].first=-1;
      else v[i].first=1;
      v[i].first += v[i-1].first;
      v[i].second=i;
    }
    sort(v.begin(),v.end());
    long long sum = 0LL;
    for(auto& p:v){
      int from = IDX;
      int to = p.second + IDX;
      while(from<=to){
        if(to%2==0||to==1){
          sum += dp[to--];
        } else {
          to/=2;
          from/=2;
        }
      }
      int idx = p.second + IDX;
      while(idx){
        dp[idx]++;
        idx/=2;
      }
    }
    if(sum<1LL*N*(N+1)/2/2){
      b = c;
    } else {
      ans = max(ans, d);
      a = c + 1;
    }
  }
  cout<<ans<<'\n';
  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>
#define IDX 131072
using namespace std;
int ans;
int N;
int m[111111],t[111111];
int main(){
  cin>>N;
  for(int i=0;i<N;i++){
    cin>>m[i];
    t[i]=m[i];
  }
  sort(t,t+N);
  int a=0,b=N;
  while(a<b){
    int c=(a+b)/2;
    int d=t[c];
    vector<pair<int,int>>v(N+1);
    vector<int>dp(IDX*2);
    for(int i=1;i<=N;i++){
      if(m[i-1]<d)v[i].first=-1;
      else v[i].first=1;
      v[i].first += v[i-1].first;
      v[i].second=i;
    }
    sort(v.begin(),v.end());
    long long sum = 0LL;
    for(auto& p:v){
      int from = IDX;
      int to = p.second + IDX;
      while(from<=to){
        if(to%2==0||to==1){
          sum += dp[to--];
        } else {
          to/=2;
          from/=2;
        }
      }
      int idx = p.second + IDX;
      while(idx){
        dp[idx]++;
        idx/=2;
      }
    }
    if(sum<1LL*N*(N+1)/2/2){
      b = c;
    } else {
      ans = max(ans, d);
      a = c + 1;
    }
  }
  cout<<ans<<'\n';
  return 0;
}
```