#include <bits/stdc++.h>
using namespace std;
ostream& operator<<(ostream& out, pair<int, int>& o) {
  out << "(" << o.first << ", " << o.second << ") ";
  return out;
}
template <class Int>
ostream& operator<<(ostream& out, vector<Int>& o) {
  for (auto it : o) out << it << " ";
  return out;
}
template <class Int>
istream& operator>>(istream& in, vector<Int>& i) {
  for (auto& it : i) in >> it;
  return in;
}
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  int n, m;
  cin >> n >> m;
  vector<int> v(n);
  cin >> v;
  map<int, int> Big;
  int root = 0;
  for (int i = 0; i < n; i++)
    if (v[i] == m) root = i;
  int cur = 0;
  Big[0] = 1;
  Big[1] = 0;
  for (int i = root + 1; i < n; i++) {
    cur += v[i] > v[root] ? 1 : -1;
    if (Big.count(cur))
      Big[cur] += 1;
    else
      Big[cur] = 1;
  }
  long long ans = Big[0] + Big[1];
  cur = 0;
  for (int i = root - 1; i >= 0; i--) {
    cur += v[i] > v[root] ? -1 : 1;
    if (Big.count(cur)) ans += Big[cur];
    if (Big.count(cur + 1)) ans += Big[cur + 1];
  }
  cout << ans << endl;
}

### Prompt

Your challenge is to write 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;
ostream& operator<<(ostream& out, pair<int, int>& o) {
  out << "(" << o.first << ", " << o.second << ") ";
  return out;
}
template <class Int>
ostream& operator<<(ostream& out, vector<Int>& o) {
  for (auto it : o) out << it << " ";
  return out;
}
template <class Int>
istream& operator>>(istream& in, vector<Int>& i) {
  for (auto& it : i) in >> it;
  return in;
}
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  int n, m;
  cin >> n >> m;
  vector<int> v(n);
  cin >> v;
  map<int, int> Big;
  int root = 0;
  for (int i = 0; i < n; i++)
    if (v[i] == m) root = i;
  int cur = 0;
  Big[0] = 1;
  Big[1] = 0;
  for (int i = root + 1; i < n; i++) {
    cur += v[i] > v[root] ? 1 : -1;
    if (Big.count(cur))
      Big[cur] += 1;
    else
      Big[cur] = 1;
  }
  long long ans = Big[0] + Big[1];
  cur = 0;
  for (int i = root - 1; i >= 0; i--) {
    cur += v[i] > v[root] ? -1 : 1;
    if (Big.count(cur)) ans += Big[cur];
    if (Big.count(cur + 1)) ans += Big[cur + 1];
  }
  cout << ans << endl;
}

```