#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long int n, m, ind = -1;
  cin >> n >> m;
  map<long long int, long long int> premp, sufmp;
  vector<long long int> a(n);
  vector<pair<long long int, long long int>> pre, suf;
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] == m) {
      ind = i;
    }
  }
  long long int ans = 1;
  pre.push_back({0, 0});
  for (long long int i = ind + 1; i < n; i++) {
    if (a[i] < m) {
      pre.push_back({pre.back().first + 1, pre.back().second});
      premp[pre.back().second - pre.back().first]++;
    } else {
      pre.push_back({pre.back().first, pre.back().second + 1});
      premp[pre.back().second - pre.back().first]++;
    }
    if (pre.back().second - pre.back().first == 1 ||
        pre.back().second - pre.back().first == 0) {
      ans++;
    }
  }
  suf.push_back({0, 0});
  for (long long int i = ind - 1; i >= 0; i--) {
    if (a[i] < m) {
      suf.push_back({suf.back().first + 1, suf.back().second});
      sufmp[suf.back().second - suf.back().first]++;
    } else {
      suf.push_back({suf.back().first, suf.back().second + 1});
      sufmp[suf.back().second - suf.back().first]++;
    }
    if (suf.back().second - suf.back().first == 1 ||
        suf.back().second - suf.back().first == 0) {
      ans++;
    }
  }
  long long int a1 = 0;
  for (long long int i = 1; i < pre.size(); i++) {
    long long int t1 = pre[i].first - pre[i].second;
    a1 += sufmp[t1];
    a1 += sufmp[t1 + 1];
  }
  for (long long int i = 1; i < suf.size(); i++) {
    long long int t1 = suf[i].first - suf[i].second;
    a1 += premp[t1];
    a1 += premp[t1 + 1];
  }
  a1 /= 2;
  ans += a1;
  cout << ans << endl;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long int n, m, ind = -1;
  cin >> n >> m;
  map<long long int, long long int> premp, sufmp;
  vector<long long int> a(n);
  vector<pair<long long int, long long int>> pre, suf;
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] == m) {
      ind = i;
    }
  }
  long long int ans = 1;
  pre.push_back({0, 0});
  for (long long int i = ind + 1; i < n; i++) {
    if (a[i] < m) {
      pre.push_back({pre.back().first + 1, pre.back().second});
      premp[pre.back().second - pre.back().first]++;
    } else {
      pre.push_back({pre.back().first, pre.back().second + 1});
      premp[pre.back().second - pre.back().first]++;
    }
    if (pre.back().second - pre.back().first == 1 ||
        pre.back().second - pre.back().first == 0) {
      ans++;
    }
  }
  suf.push_back({0, 0});
  for (long long int i = ind - 1; i >= 0; i--) {
    if (a[i] < m) {
      suf.push_back({suf.back().first + 1, suf.back().second});
      sufmp[suf.back().second - suf.back().first]++;
    } else {
      suf.push_back({suf.back().first, suf.back().second + 1});
      sufmp[suf.back().second - suf.back().first]++;
    }
    if (suf.back().second - suf.back().first == 1 ||
        suf.back().second - suf.back().first == 0) {
      ans++;
    }
  }
  long long int a1 = 0;
  for (long long int i = 1; i < pre.size(); i++) {
    long long int t1 = pre[i].first - pre[i].second;
    a1 += sufmp[t1];
    a1 += sufmp[t1 + 1];
  }
  for (long long int i = 1; i < suf.size(); i++) {
    long long int t1 = suf[i].first - suf[i].second;
    a1 += premp[t1];
    a1 += premp[t1 + 1];
  }
  a1 /= 2;
  ans += a1;
  cout << ans << endl;
}

```