#include <bits/stdc++.h>
using namespace std;
int64_t* a;
int64_t n;
int64_t getCount(int64_t m) {
  int64_t* b = new int64_t[2 * n + 10];
  memset(b, 0, 2 * (n + 5) * sizeof(int64_t));
  b[n + 1] = 1;
  int64_t now = 0;
  int64_t ans = 0;
  int64_t sum = 0;
  for (int64_t i = 1; i <= n; i++) {
    if (a[i] < m) {
      sum -= b[n + now--];
    }
    if (a[i] >= m) {
      sum += b[n + ++now];
    }
    ans += sum;
    b[n + now + 1LL]++;
  }
  return ans;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int64_t m;
  cin >> n >> m;
  a = new int64_t[n + 1];
  for (int64_t i = 1; i <= n; i++) cin >> a[i];
  cout << getCount(m) - getCount(m + 1) << endl;
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution to the following problem:

You are given an integer sequence a_1, a_2, ..., a_n.

Find the number of pairs of indices (l, r) (1 ≤ l ≤ r ≤ n) such that the value of median of a_l, a_{l+1}, ..., a_r is exactly the given number m.

The median of a sequence is the value of an element which is in the middle of the sequence after sorting it in non-decreasing order. If the length of the sequence is even, the left of two middle elements is used.

For example, if a=[4, 2, 7, 5] then its median is 4 since after sorting the sequence, it will look like [2, 4, 5, 7] and the left of two middle elements is equal to 4. The median of [7, 1, 2, 9, 6] equals 6 since after sorting, the value 6 will be in the middle of the sequence.

Write a program to find the number of pairs of indices (l, r) (1 ≤ l ≤ r ≤ n) such that the value of median of a_l, a_{l+1}, ..., a_r is exactly the given number m.

Input

The first line contains integers n and m (1 ≤ n,m ≤ 2⋅10^5) — the length of the given sequence and the required value of the median.

The second line contains an integer sequence a_1, a_2, ..., a_n (1 ≤ a_i ≤ 2⋅10^5).

Output

Print the required number.

Examples

Input

5 4
1 4 5 60 4


Output

8


Input

3 1
1 1 1


Output

6


Input

15 2
1 2 3 1 2 3 1 2 3 1 2 3 1 2 3


Output

97

Note

In the first example, the suitable pairs of indices are: (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 5), (4, 5) and (5, 5).

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int64_t* a;
int64_t n;
int64_t getCount(int64_t m) {
  int64_t* b = new int64_t[2 * n + 10];
  memset(b, 0, 2 * (n + 5) * sizeof(int64_t));
  b[n + 1] = 1;
  int64_t now = 0;
  int64_t ans = 0;
  int64_t sum = 0;
  for (int64_t i = 1; i <= n; i++) {
    if (a[i] < m) {
      sum -= b[n + now--];
    }
    if (a[i] >= m) {
      sum += b[n + ++now];
    }
    ans += sum;
    b[n + now + 1LL]++;
  }
  return ans;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int64_t m;
  cin >> n >> m;
  a = new int64_t[n + 1];
  for (int64_t i = 1; i <= n; i++) cin >> a[i];
  cout << getCount(m) - getCount(m + 1) << endl;
  return 0;
}

```