#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, N) for (int i = 0; i < (int)N; i++)

ll N;
vector<int> S, a;

ll merge(int l, int m, int r) {
  int i, j, k;
  ll cnt = 0;
  int n1 = m - l;
  int n2 = r - m;
  vector<int> L(n1+1), R(n2+1);
  rep(i,n1) L[i] = S[l+i];
  rep(i,n2) R[i] = S[m+i];
  L[n1] = R[n2] = 1e9+1;
  i = j = 0;
  for(k = l; k < r; k++) {
    if(L[i] <= R[j]) {
      S[k] = L[i++];
      cnt += n2 - j;
    } else {
      S[k] = R[j++];
    }
  }
  return cnt;
}

ll count(int l, int r) {
  if(r - l <= 1) return 0;

  int m;
  ll v1, v2, v3;
  m = (l + r) / 2;
  v1 = count(l, m);
  v2 = count(m, r);
  v3 = merge(l, m, r);
  return v1 + v2 + v3;
}

int main () {
  cin >> N;
  a = vector<int>(N);
  S = vector<int>(N+1);
  rep(i,N) cin >> a[i];

  // [l, r)
  int l = 0, r = 1e9 + 1;
  ll t = (N*(N+1)/2 + 1)/2;
  while(r - l > 1) {
    int x = (l + r) / 2;
    S[0] = 0;
    rep(i,N) S[i+1] = S[i] + (a[i] >= x ? 1 : -1);
    if(count(0, N+1) >= t) {
      l = x;
    } else {
      r = x;
    }
  }

  cout << l << 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>
using namespace std;
typedef long long ll;
#define rep(i, N) for (int i = 0; i < (int)N; i++)

ll N;
vector<int> S, a;

ll merge(int l, int m, int r) {
  int i, j, k;
  ll cnt = 0;
  int n1 = m - l;
  int n2 = r - m;
  vector<int> L(n1+1), R(n2+1);
  rep(i,n1) L[i] = S[l+i];
  rep(i,n2) R[i] = S[m+i];
  L[n1] = R[n2] = 1e9+1;
  i = j = 0;
  for(k = l; k < r; k++) {
    if(L[i] <= R[j]) {
      S[k] = L[i++];
      cnt += n2 - j;
    } else {
      S[k] = R[j++];
    }
  }
  return cnt;
}

ll count(int l, int r) {
  if(r - l <= 1) return 0;

  int m;
  ll v1, v2, v3;
  m = (l + r) / 2;
  v1 = count(l, m);
  v2 = count(m, r);
  v3 = merge(l, m, r);
  return v1 + v2 + v3;
}

int main () {
  cin >> N;
  a = vector<int>(N);
  S = vector<int>(N+1);
  rep(i,N) cin >> a[i];

  // [l, r)
  int l = 0, r = 1e9 + 1;
  ll t = (N*(N+1)/2 + 1)/2;
  while(r - l > 1) {
    int x = (l + r) / 2;
    S[0] = 0;
    rep(i,N) S[i+1] = S[i] + (a[i] >= x ? 1 : -1);
    if(count(0, N+1) >= t) {
      l = x;
    } else {
      r = x;
    }
  }

  cout << l << endl;

  return 0;
}

```