#include <bits/stdc++.h>
#define F first
#define S second
using namespace std;
using ll = long long;
typedef pair<int, int> P;
ll Mod = 1000000007;
int main() {
  ll N;
  cin >> N;
  ll a[N];
  for (int i = 0; i < N; i++) {
    cin >> a[i];
  }
  ll m = (N * (N+1LL)) / 2LL;
  m += 1LL;
  m /= 2LL;
  ll left = -1;
  ll right = Mod * 2;
  while (right - left > 1LL) {
    ll mid = (left + right) / 2LL;
    ll R[N+1];
    // 450個ごと　平方分割
    // -200000 ~ 200000
    vector<ll> num(400001);
    vector<ll> B(888);
    ll sum = 0;
    R[0] = 0;
    for (int i = 0; i < N; i++) {
      if (a[i] >= mid) {
        R[i+1] = R[i] + 1;
      } else {
        R[i+1] = R[i] - 1; 
      }
    }
    num[200000]++;
    B[200000/450]++;
    
    for (ll i = 0; i < N; i++) {
      ll l = (R[i+1] + 200000) / 450;
      ll s = 0;
      for (ll j = 0; j < l; j++) {
        s += B[j];
      }
      for (ll j = l * 450; j <= (R[i+1] + 200000); j++) {
        s += num[j];
      }
      sum += s;
      num[R[i+1]+200000]++;
      B[(R[i+1]+200000)/450]++;
    }
    if (sum >= m) {
      left = mid;
    } else {
      right = mid;
    }
  }
  cout << left << endl;
  return 0;
}
### 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 <bits/stdc++.h>
#define F first
#define S second
using namespace std;
using ll = long long;
typedef pair<int, int> P;
ll Mod = 1000000007;
int main() {
  ll N;
  cin >> N;
  ll a[N];
  for (int i = 0; i < N; i++) {
    cin >> a[i];
  }
  ll m = (N * (N+1LL)) / 2LL;
  m += 1LL;
  m /= 2LL;
  ll left = -1;
  ll right = Mod * 2;
  while (right - left > 1LL) {
    ll mid = (left + right) / 2LL;
    ll R[N+1];
    // 450個ごと　平方分割
    // -200000 ~ 200000
    vector<ll> num(400001);
    vector<ll> B(888);
    ll sum = 0;
    R[0] = 0;
    for (int i = 0; i < N; i++) {
      if (a[i] >= mid) {
        R[i+1] = R[i] + 1;
      } else {
        R[i+1] = R[i] - 1; 
      }
    }
    num[200000]++;
    B[200000/450]++;
    
    for (ll i = 0; i < N; i++) {
      ll l = (R[i+1] + 200000) / 450;
      ll s = 0;
      for (ll j = 0; j < l; j++) {
        s += B[j];
      }
      for (ll j = l * 450; j <= (R[i+1] + 200000); j++) {
        s += num[j];
      }
      sum += s;
      num[R[i+1]+200000]++;
      B[(R[i+1]+200000)/450]++;
    }
    if (sum >= m) {
      left = mid;
    } else {
      right = mid;
    }
  }
  cout << left << endl;
  return 0;
}
```