#include <bits/stdc++.h>
using namespace std;
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << "[";
  typename vector<T>::const_iterator it;
  for (it = v.begin(); it != v.end(); it++) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &v) {
  os << "{";
  typename set<T>::const_iterator it;
  for (it = v.begin(); it != v.end(); it++) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "}";
}
template <typename A, typename B>
ostream &operator<<(ostream &os, const map<A, B> &v) {
  os << "[";
  typename map<A, B>::const_iterator it;
  for (it = v.begin(); it != v.end(); it++) {
    if (it != v.begin()) os << ", ";
    os << it->first << " -> " << it->second;
  }
  return os << "]";
}
template <typename T>
inline void printV(vector<T> &a, long long n) {
  for (long long i = 0; i < n; i++) {
    cerr << a[i] << " ";
  }
  cerr << "\n";
}
template <typename T>
inline void printVV(vector<vector<T>> &a) {
  for (long long i = 0; i < ((long long)((a).size())); i++) {
    cerr << a[i] << "\n";
  }
  cerr << "\n";
}
template <typename T>
inline void clearVV(vector<vector<T>> &a) {
  for (auto &x : a) {
    x.clear();
  }
  a.clear();
}
template <typename T>
inline void assignVV(vector<vector<T>> &a, T val) {
  for (long long i = 0; i < ((long long)((a).size())); i++) {
    a[i].assign(((long long)((a[i]).size())), val);
  }
}
template <typename T>
T gcd(T a, T b) {
  if (a == 0 or b == 0) return a + b;
  return gcd(b, a % b);
}
template <typename T>
T lcm(T a, T b) {
  if (a == 0 or b == 0) return 0;
  return a / gcd(a, b) * b;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long double eps = LDBL_EPSILON;
const long long inf = 5e8;
const long long modn = 1e9 + 7;
const long long MAX = 1e5;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, m;
  cin >> n >> m;
  m--;
  vector<long long> a(n), pos(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    a[i]--;
    pos[a[i]] = i;
  }
  vector<long long> less(n), great(n);
  for (long long i = pos[m] - 1; i >= 0; i--) {
    if (a[i] < m) {
      less[i] = less[i + 1] + 1;
      great[i] = great[i + 1];
    } else {
      less[i] = less[i + 1];
      great[i] = great[i + 1] + 1;
    }
  }
  for (long long i = pos[m] + 1; i < n; i++) {
    if (a[i] < m) {
      less[i] = less[i - 1] + 1;
      great[i] = great[i - 1];
    } else {
      less[i] = less[i - 1];
      great[i] = great[i - 1] + 1;
    }
  }
  map<long long, long long> mm;
  for (long long i = pos[m]; i >= 0; i--) {
    mm[less[i] - great[i]]++;
  }
  long long ans = 0;
  for (long long i = pos[m]; i < n; i++) {
    long long dif = great[i] - less[i];
    ans += mm[dif];
    ans += mm[dif - 1];
  }
  cout << ans << "\n";
  return 0;
}

### Prompt

Your task is to create 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 <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << "[";
  typename vector<T>::const_iterator it;
  for (it = v.begin(); it != v.end(); it++) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &v) {
  os << "{";
  typename set<T>::const_iterator it;
  for (it = v.begin(); it != v.end(); it++) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "}";
}
template <typename A, typename B>
ostream &operator<<(ostream &os, const map<A, B> &v) {
  os << "[";
  typename map<A, B>::const_iterator it;
  for (it = v.begin(); it != v.end(); it++) {
    if (it != v.begin()) os << ", ";
    os << it->first << " -> " << it->second;
  }
  return os << "]";
}
template <typename T>
inline void printV(vector<T> &a, long long n) {
  for (long long i = 0; i < n; i++) {
    cerr << a[i] << " ";
  }
  cerr << "\n";
}
template <typename T>
inline void printVV(vector<vector<T>> &a) {
  for (long long i = 0; i < ((long long)((a).size())); i++) {
    cerr << a[i] << "\n";
  }
  cerr << "\n";
}
template <typename T>
inline void clearVV(vector<vector<T>> &a) {
  for (auto &x : a) {
    x.clear();
  }
  a.clear();
}
template <typename T>
inline void assignVV(vector<vector<T>> &a, T val) {
  for (long long i = 0; i < ((long long)((a).size())); i++) {
    a[i].assign(((long long)((a[i]).size())), val);
  }
}
template <typename T>
T gcd(T a, T b) {
  if (a == 0 or b == 0) return a + b;
  return gcd(b, a % b);
}
template <typename T>
T lcm(T a, T b) {
  if (a == 0 or b == 0) return 0;
  return a / gcd(a, b) * b;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long double eps = LDBL_EPSILON;
const long long inf = 5e8;
const long long modn = 1e9 + 7;
const long long MAX = 1e5;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, m;
  cin >> n >> m;
  m--;
  vector<long long> a(n), pos(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    a[i]--;
    pos[a[i]] = i;
  }
  vector<long long> less(n), great(n);
  for (long long i = pos[m] - 1; i >= 0; i--) {
    if (a[i] < m) {
      less[i] = less[i + 1] + 1;
      great[i] = great[i + 1];
    } else {
      less[i] = less[i + 1];
      great[i] = great[i + 1] + 1;
    }
  }
  for (long long i = pos[m] + 1; i < n; i++) {
    if (a[i] < m) {
      less[i] = less[i - 1] + 1;
      great[i] = great[i - 1];
    } else {
      less[i] = less[i - 1];
      great[i] = great[i - 1] + 1;
    }
  }
  map<long long, long long> mm;
  for (long long i = pos[m]; i >= 0; i--) {
    mm[less[i] - great[i]]++;
  }
  long long ans = 0;
  for (long long i = pos[m]; i < n; i++) {
    long long dif = great[i] - less[i];
    ans += mm[dif];
    ans += mm[dif - 1];
  }
  cout << ans << "\n";
  return 0;
}

```