#include <bits/stdc++.h>
using namespace std;
pair<int, int> t[4] = {make_pair(0, 1), make_pair(1, 0), make_pair(-1, 0),
                       make_pair(0, -1)};
class compare {
 public:
  bool operator()(pair<int, int> a, pair<int, int> b) {
    return (a.second > b.second);
  }
};
bool compare2(long long int a, long long int b) { return (a < b); }
int main() {
  int n, m1;
  cin >> n >> m1;
  int a[n + 1], k = -1;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] == m1) {
      k = i;
    }
  }
  long long int count = 0;
  if (k == -1) {
    cout << 0 << endl;
    ;
  } else {
    long long int lessm = 0, greatm = 0;
    map<long long int, long long int> m;
    for (int i = k; i < n; i++) {
      if (a[i] <= m1)
        lessm++;
      else
        greatm++;
      long long int diff = lessm - greatm;
      m[diff]++;
      if (diff >= 0 && diff <= 1) count++;
    }
    lessm = 0;
    greatm = 0;
    for (int i = k - 1; i >= 0; i--) {
      if (a[i] <= m1)
        lessm++;
      else
        greatm++;
      long long int diff = lessm - greatm;
      long long int p = -1 * diff;
      if (p < 0) {
        count += m[p];
        count += m[p + 1];
      } else if (p > 0) {
        count += m[p];
        count += m[p + 1];
      } else {
        count += m[p];
        count += m[p + 1];
      }
    }
    cout << count << endl;
    ;
  }
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
pair<int, int> t[4] = {make_pair(0, 1), make_pair(1, 0), make_pair(-1, 0),
                       make_pair(0, -1)};
class compare {
 public:
  bool operator()(pair<int, int> a, pair<int, int> b) {
    return (a.second > b.second);
  }
};
bool compare2(long long int a, long long int b) { return (a < b); }
int main() {
  int n, m1;
  cin >> n >> m1;
  int a[n + 1], k = -1;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] == m1) {
      k = i;
    }
  }
  long long int count = 0;
  if (k == -1) {
    cout << 0 << endl;
    ;
  } else {
    long long int lessm = 0, greatm = 0;
    map<long long int, long long int> m;
    for (int i = k; i < n; i++) {
      if (a[i] <= m1)
        lessm++;
      else
        greatm++;
      long long int diff = lessm - greatm;
      m[diff]++;
      if (diff >= 0 && diff <= 1) count++;
    }
    lessm = 0;
    greatm = 0;
    for (int i = k - 1; i >= 0; i--) {
      if (a[i] <= m1)
        lessm++;
      else
        greatm++;
      long long int diff = lessm - greatm;
      long long int p = -1 * diff;
      if (p < 0) {
        count += m[p];
        count += m[p + 1];
      } else if (p > 0) {
        count += m[p];
        count += m[p + 1];
      } else {
        count += m[p];
        count += m[p + 1];
      }
    }
    cout << count << endl;
    ;
  }
}

```