#include <bits/stdc++.h>
using namespace std;
int p[200005];
int d[200005];
int main() {
  memset(p, 0, sizeof(p));
  memset(d, 0, sizeof(d));
  int n, m;
  int ii;
  scanf("%d %d", &n, &m);
  unordered_map<int, int> t1(100000), t2(100000);
  int ds = 0;
  int pds = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &p[i]);
    if (p[i] == m) {
      ii = i;
    }
  }
  for (int i = 0; i < n; i++) {
    if (p[i] < m) {
      d[i] = 1;
    } else if (p[i] > m) {
      d[i] = -1;
    } else {
      d[i] = 0;
    }
    if (i >= ii) {
      int l = i - ii + 1;
      ds += d[i];
      if (l & 1) {
        t1[ds]++;
      } else {
        t2[ds]++;
      }
    } else {
      pds += d[i];
    }
  }
  long long ans = 0;
  for (int i = 0; i <= ii; i++) {
    int l1 = ii - i;
    if (l1 & 1) {
      ans += t1[-1 - pds];
      ans += t2[-pds];
    } else {
      ans += t1[-pds];
      ans += t2[-1 - pds];
    }
    pds -= d[i];
  }
  printf("%lld\n", ans);
  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 p[200005];
int d[200005];
int main() {
  memset(p, 0, sizeof(p));
  memset(d, 0, sizeof(d));
  int n, m;
  int ii;
  scanf("%d %d", &n, &m);
  unordered_map<int, int> t1(100000), t2(100000);
  int ds = 0;
  int pds = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &p[i]);
    if (p[i] == m) {
      ii = i;
    }
  }
  for (int i = 0; i < n; i++) {
    if (p[i] < m) {
      d[i] = 1;
    } else if (p[i] > m) {
      d[i] = -1;
    } else {
      d[i] = 0;
    }
    if (i >= ii) {
      int l = i - ii + 1;
      ds += d[i];
      if (l & 1) {
        t1[ds]++;
      } else {
        t2[ds]++;
      }
    } else {
      pds += d[i];
    }
  }
  long long ans = 0;
  for (int i = 0; i <= ii; i++) {
    int l1 = ii - i;
    if (l1 & 1) {
      ans += t1[-1 - pds];
      ans += t2[-pds];
    } else {
      ans += t1[-pds];
      ans += t2[-1 - pds];
    }
    pds -= d[i];
  }
  printf("%lld\n", ans);
  return (0);
}

```