#include <bits/stdc++.h>
using namespace std;
template <class T>
ostream &operator<<(ostream &os, const pair<T, T> &p) {
  return os << "(" << p.first << "," << p.second << ")";
}
template <class T>
ostream &operator<<(ostream &os, const pair<const T, T> &p) {
  return os << "(" << p.first << "," << p.second << ")";
}
const double PI = acos(-1);
const int INF = 0x3f3f3f3f;
const int SZ = 2e5 + 10;
const long long int mod = 1e9 + 7;
int main() {
  int n, m;
  scanf(" %d %d", &n, &m);
  vector<int> v;
  int id;
  for (int i = 0; i != n; 0 < n ? i++ : i--) {
    int x;
    scanf(" %d", &x);
    if (x == m) id = v.size();
    v.push_back(x);
  }
  map<int, int> esq, dir;
  int maiores = 0, menores = 0;
  for (int i = id - 1; i >= 0; i--) {
    if (v[i] < m)
      menores++;
    else
      maiores++;
    int dist = maiores - menores;
    esq[dist]++;
  }
  maiores = 0, menores = 0;
  for (int i = id + 1; i < n; i++) {
    if (v[i] < m)
      menores++;
    else
      maiores++;
    int dist = maiores - menores;
    dir[dist]++;
  }
  long long int res = esq[0] + esq[1] + dir[0] + dir[1] + 1;
  for (auto p : esq) {
    res += p.second * 1LL * dir[-1 * p.first];
    res += p.second * 1LL * dir[-1 * p.first + 1];
  }
  cout << res << "\n";
  return 0;
}

### Prompt

Please formulate 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;
template <class T>
ostream &operator<<(ostream &os, const pair<T, T> &p) {
  return os << "(" << p.first << "," << p.second << ")";
}
template <class T>
ostream &operator<<(ostream &os, const pair<const T, T> &p) {
  return os << "(" << p.first << "," << p.second << ")";
}
const double PI = acos(-1);
const int INF = 0x3f3f3f3f;
const int SZ = 2e5 + 10;
const long long int mod = 1e9 + 7;
int main() {
  int n, m;
  scanf(" %d %d", &n, &m);
  vector<int> v;
  int id;
  for (int i = 0; i != n; 0 < n ? i++ : i--) {
    int x;
    scanf(" %d", &x);
    if (x == m) id = v.size();
    v.push_back(x);
  }
  map<int, int> esq, dir;
  int maiores = 0, menores = 0;
  for (int i = id - 1; i >= 0; i--) {
    if (v[i] < m)
      menores++;
    else
      maiores++;
    int dist = maiores - menores;
    esq[dist]++;
  }
  maiores = 0, menores = 0;
  for (int i = id + 1; i < n; i++) {
    if (v[i] < m)
      menores++;
    else
      maiores++;
    int dist = maiores - menores;
    dir[dist]++;
  }
  long long int res = esq[0] + esq[1] + dir[0] + dir[1] + 1;
  for (auto p : esq) {
    res += p.second * 1LL * dir[-1 * p.first];
    res += p.second * 1LL * dir[-1 * p.first + 1];
  }
  cout << res << "\n";
  return 0;
}

```