#include<iostream>
using namespace std;

const int MAXN = 100100;

int n;
int a[MAXN], p[MAXN], csum[2*MAXN];

bool good(int c){
  for(int i=0;i<n;i++){
    if (a[i] >= c) p[i] = 1;
    else p[i] = -1;
  }
  for(int i=0;i<2*MAXN;i++) csum[i] = 0;

  int cs = MAXN;
  csum[cs] = 1;
  int ctot = 1;
  long long res = 0;
  for(int i=0;i<n;i++){
    if(p[i] == -1){
      ctot -= csum[cs];
      cs--;
    }else{
      ctot += csum[cs+1];
      cs++;
    }
    res += ctot;
    csum[cs]++;
    ctot++;
  }
  return res * 2 >= (n*((long long)n+1))/2;
}

int main(){
  cin >> n;
  for(int i=0;i<n;i++) cin >> a[i];

  int l=0, r=1e9;
  while(l<r){
    int mid = (l+r+1)/2;
    if(good(mid)) l = mid;
    else r = mid - 1;
  }
  cout << l << endl;
}

### Prompt

In Cpp, your task is to solve 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<iostream>
using namespace std;

const int MAXN = 100100;

int n;
int a[MAXN], p[MAXN], csum[2*MAXN];

bool good(int c){
  for(int i=0;i<n;i++){
    if (a[i] >= c) p[i] = 1;
    else p[i] = -1;
  }
  for(int i=0;i<2*MAXN;i++) csum[i] = 0;

  int cs = MAXN;
  csum[cs] = 1;
  int ctot = 1;
  long long res = 0;
  for(int i=0;i<n;i++){
    if(p[i] == -1){
      ctot -= csum[cs];
      cs--;
    }else{
      ctot += csum[cs+1];
      cs++;
    }
    res += ctot;
    csum[cs]++;
    ctot++;
  }
  return res * 2 >= (n*((long long)n+1))/2;
}

int main(){
  cin >> n;
  for(int i=0;i<n;i++) cin >> a[i];

  int l=0, r=1e9;
  while(l<r){
    int mid = (l+r+1)/2;
    if(good(mid)) l = mid;
    else r = mid - 1;
  }
  cout << l << endl;
}

```