#include <bits/stdc++.h>
using namespace std;
typedef vector<int> vtype;
int Mod(int& x, int& m) {
  int r = x % m;
  return r < 0 ? r + m : r;
}
int arr[200005];
map<long long int, long long int> below;
map<long long int, long long int> above;
int main() {
  {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
  }
  int n, m;
  cin >> n >> m;
  int midx;
  for (int i = 1; i <= n; i++) {
    cin >> arr[i];
    if (arr[i] == m) midx = i;
  }
  int highs = 0;
  int lows = 0;
  for (int i = midx; i > 0; i--) {
    if (arr[i] > m)
      highs++;
    else if (arr[i] < m)
      lows++;
    if (below.count(highs - lows))
      below[highs - lows]++;
    else
      below[highs - lows] = 1;
  }
  highs = 0;
  lows = 0;
  if (midx != n) {
    for (int i = midx + 1; i <= n; i++) {
      if (arr[i] > m)
        highs++;
      else if (arr[i] < m)
        lows++;
      if (above.count(highs - lows))
        above[highs - lows]++;
      else
        above[highs - lows] = 1;
    }
  }
  long long int ans = 0;
  long long int cnt;
  int diff;
  for (pair<int, int> i : below) {
    diff = i.first;
    cnt = i.second;
    ans += (above.count(0 - diff) ? cnt * above[0 - diff] : 0LL) +
           (above.count(1 - diff) ? cnt * above[1 - diff] : 0LL);
  }
  cout << ans + below[0] + below[1];
  return 0;
}

### Prompt

Please create a solution in cpp 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;
typedef vector<int> vtype;
int Mod(int& x, int& m) {
  int r = x % m;
  return r < 0 ? r + m : r;
}
int arr[200005];
map<long long int, long long int> below;
map<long long int, long long int> above;
int main() {
  {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
  }
  int n, m;
  cin >> n >> m;
  int midx;
  for (int i = 1; i <= n; i++) {
    cin >> arr[i];
    if (arr[i] == m) midx = i;
  }
  int highs = 0;
  int lows = 0;
  for (int i = midx; i > 0; i--) {
    if (arr[i] > m)
      highs++;
    else if (arr[i] < m)
      lows++;
    if (below.count(highs - lows))
      below[highs - lows]++;
    else
      below[highs - lows] = 1;
  }
  highs = 0;
  lows = 0;
  if (midx != n) {
    for (int i = midx + 1; i <= n; i++) {
      if (arr[i] > m)
        highs++;
      else if (arr[i] < m)
        lows++;
      if (above.count(highs - lows))
        above[highs - lows]++;
      else
        above[highs - lows] = 1;
    }
  }
  long long int ans = 0;
  long long int cnt;
  int diff;
  for (pair<int, int> i : below) {
    diff = i.first;
    cnt = i.second;
    ans += (above.count(0 - diff) ? cnt * above[0 - diff] : 0LL) +
           (above.count(1 - diff) ? cnt * above[1 - diff] : 0LL);
  }
  cout << ans + below[0] + below[1];
  return 0;
}

```