#include <bits/stdc++.h>
using namespace std;
int a[200009];
int main() {
  int n, m, ind;
  long long int ans = 0;
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] == m) {
      ind = i;
    }
  }
  vector<int> l, r1, r2;
  l.push_back(0);
  r1.push_back(0);
  r2.push_back(-1);
  for (int i = ind - 1; i >= 0; i--) {
    if (a[i] < m) {
      l.push_back(l[l.size() - 1] + 1);
    } else {
      l.push_back(l[l.size() - 1] - 1);
    }
  }
  for (int i = ind + 1; i < n; i++) {
    if (a[i] < m) {
      r1.push_back(r1[r1.size() - 1] - 1);
      r2.push_back(r2[r2.size() - 1] - 1);
    } else {
      r1.push_back(r1[r1.size() - 1] + 1);
      r2.push_back(r2[r2.size() - 1] + 1);
    }
  }
  sort(l.begin(), l.end());
  sort(r1.begin(), r1.end());
  sort(r2.begin(), r2.end());
  for (int i = 0; i < l.size();) {
    long long int x =
        int(upper_bound(l.begin(), l.end(), l[i]) - l.begin()) - i;
    long long int y = upper_bound(r1.begin(), r1.end(), l[i]) -
                      lower_bound(r1.begin(), r1.end(), l[i]);
    long long int z = upper_bound(r2.begin(), r2.end(), l[i]) -
                      lower_bound(r2.begin(), r2.end(), l[i]);
    ans += x * y + x * z;
    i += x;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Generate a CPP solution to the following problem:

You are given a permutation p_1, p_2, ..., p_n. A permutation of length n is a sequence such that each integer between 1 and n occurs exactly once in the sequence.

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

The median of a sequence is the value of the 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 the median of p_l, p_{l+1}, ..., p_r is exactly the given number m.

Input

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

The second line contains a permutation p_1, p_2, ..., p_n (1 ≤ p_i ≤ n). Each integer between 1 and n occurs in p exactly once.

Output

Print the required number.

Examples

Input

5 4
2 4 5 3 1


Output

4


Input

5 5
1 2 3 4 5


Output

1


Input

15 8
1 15 2 14 3 13 4 8 12 5 11 6 10 7 9


Output

48

Note

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int a[200009];
int main() {
  int n, m, ind;
  long long int ans = 0;
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] == m) {
      ind = i;
    }
  }
  vector<int> l, r1, r2;
  l.push_back(0);
  r1.push_back(0);
  r2.push_back(-1);
  for (int i = ind - 1; i >= 0; i--) {
    if (a[i] < m) {
      l.push_back(l[l.size() - 1] + 1);
    } else {
      l.push_back(l[l.size() - 1] - 1);
    }
  }
  for (int i = ind + 1; i < n; i++) {
    if (a[i] < m) {
      r1.push_back(r1[r1.size() - 1] - 1);
      r2.push_back(r2[r2.size() - 1] - 1);
    } else {
      r1.push_back(r1[r1.size() - 1] + 1);
      r2.push_back(r2[r2.size() - 1] + 1);
    }
  }
  sort(l.begin(), l.end());
  sort(r1.begin(), r1.end());
  sort(r2.begin(), r2.end());
  for (int i = 0; i < l.size();) {
    long long int x =
        int(upper_bound(l.begin(), l.end(), l[i]) - l.begin()) - i;
    long long int y = upper_bound(r1.begin(), r1.end(), l[i]) -
                      lower_bound(r1.begin(), r1.end(), l[i]);
    long long int z = upper_bound(r2.begin(), r2.end(), l[i]) -
                      lower_bound(r2.begin(), r2.end(), l[i]);
    ans += x * y + x * z;
    i += x;
  }
  cout << ans << endl;
  return 0;
}

```