#include <bits/stdc++.h>
using namespace std;
int A[200006], p1[200006], p2[200006], pos[200006], neg[200006];
int main() {
  int N, i, j, k, M;
  cin >> N >> M;
  int idx;
  for (i = 1; i <= N; i++) {
    cin >> A[i];
    if (A[i] == M) idx = i;
  }
  for (i = idx; i <= N; i++) {
    if (A[i] > M)
      p2[i]++;
    else if (A[i] < M)
      p1[i]++;
    p1[i] += p1[i - 1];
    p2[i] += p2[i - 1];
    if (p2[i] > p1[i])
      pos[p2[i] - p1[i]]++;
    else if (p2[i] < p1[i])
      neg[p1[i] - p2[i]]++;
    else {
      pos[0]++;
      neg[0]++;
    }
  }
  for (i = 1; i <= idx; ++i) {
    if (A[i] > M) p2[i]++;
    if (A[i] < M) p1[i]++;
    p1[i] += p1[i - 1];
    p2[i] += p2[i - 1];
  }
  long long ans = 0;
  for (i = 1; i <= idx; i++) {
    int bada = p2[idx] - p2[i - 1];
    int chota = p1[idx] - p1[i - 1];
    if (bada > chota) {
      ans += (long long)neg[bada - chota];
      ans += (long long)neg[bada - chota - 1];
    } else if (chota > bada) {
      ans += (long long)pos[chota - bada];
      ans += (long long)pos[chota - bada + 1];
    } else {
      ans += (long long)pos[0];
      ans += (long long)pos[1];
    }
  }
  cout << ans << "\n";
  return 0;
}

### Prompt

Develop a solution in Cpp 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;
int A[200006], p1[200006], p2[200006], pos[200006], neg[200006];
int main() {
  int N, i, j, k, M;
  cin >> N >> M;
  int idx;
  for (i = 1; i <= N; i++) {
    cin >> A[i];
    if (A[i] == M) idx = i;
  }
  for (i = idx; i <= N; i++) {
    if (A[i] > M)
      p2[i]++;
    else if (A[i] < M)
      p1[i]++;
    p1[i] += p1[i - 1];
    p2[i] += p2[i - 1];
    if (p2[i] > p1[i])
      pos[p2[i] - p1[i]]++;
    else if (p2[i] < p1[i])
      neg[p1[i] - p2[i]]++;
    else {
      pos[0]++;
      neg[0]++;
    }
  }
  for (i = 1; i <= idx; ++i) {
    if (A[i] > M) p2[i]++;
    if (A[i] < M) p1[i]++;
    p1[i] += p1[i - 1];
    p2[i] += p2[i - 1];
  }
  long long ans = 0;
  for (i = 1; i <= idx; i++) {
    int bada = p2[idx] - p2[i - 1];
    int chota = p1[idx] - p1[i - 1];
    if (bada > chota) {
      ans += (long long)neg[bada - chota];
      ans += (long long)neg[bada - chota - 1];
    } else if (chota > bada) {
      ans += (long long)pos[chota - bada];
      ans += (long long)pos[chota - bada + 1];
    } else {
      ans += (long long)pos[0];
      ans += (long long)pos[1];
    }
  }
  cout << ans << "\n";
  return 0;
}

```