#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("Ofast")
template <typename T>
using v = vector<T>;
const long long inf = 4611686018427387903;
void read() {}
template <typename Arg, typename... Args>
void read(Arg& arg, Args&... args) {
  cin >> (arg);
  read(args...);
}
void print() {}
template <typename Arg, typename... Args>
void print(Arg arg, Args... args) {
  cout << (arg) << " ";
  print(args...);
}
void debug() { cout << endl; }
template <typename Arg, typename... Args>
void debug(Arg arg, Args... args) {
  cout << (arg) << " ";
  debug(args...);
}
long long getInt() {
  long long a;
  cin >> a;
  return a;
}
void run() {
  long long n, m;
  read(n, m);
  long long a[n];
  for (long long i = (0); i < (n); i++) cin >> a[i];
  long long lessM[n], equalM[n], moreM[n];
  lessM[0] = a[0] < m;
  equalM[0] = a[0] == m;
  moreM[0] = a[0] > m;
  for (long long i = (1); i < (n); i++) {
    lessM[i] = (a[i] < m) + lessM[i - 1];
    equalM[i] = (a[i] == m) + equalM[i - 1];
    moreM[i] = (a[i] > m) + moreM[i - 1];
  }
  long long penalty[n];
  for (long long i = (0); i < (n); i++) penalty[i] = moreM[i] - lessM[i];
  long long ans = 0;
  long long nearestM = -1;
  map<long long, long long> pos;
  pos[0] = 1;
  for (long long r = (0); r < (n); r++) {
    if (a[r] == m) {
      for (long long i = (nearestM + (nearestM == -1)); i < (r); i++)
        pos[penalty[i]]++;
      nearestM = r;
    }
    if (nearestM == -1) continue;
    ans += pos[penalty[r]] + pos[penalty[r] - 1];
  }
  cout << ans;
}
int32_t main() {
  srand(time(0));
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed;
  cout << setprecision(15);
  run();
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
#pragma GCC optimize("Ofast")
template <typename T>
using v = vector<T>;
const long long inf = 4611686018427387903;
void read() {}
template <typename Arg, typename... Args>
void read(Arg& arg, Args&... args) {
  cin >> (arg);
  read(args...);
}
void print() {}
template <typename Arg, typename... Args>
void print(Arg arg, Args... args) {
  cout << (arg) << " ";
  print(args...);
}
void debug() { cout << endl; }
template <typename Arg, typename... Args>
void debug(Arg arg, Args... args) {
  cout << (arg) << " ";
  debug(args...);
}
long long getInt() {
  long long a;
  cin >> a;
  return a;
}
void run() {
  long long n, m;
  read(n, m);
  long long a[n];
  for (long long i = (0); i < (n); i++) cin >> a[i];
  long long lessM[n], equalM[n], moreM[n];
  lessM[0] = a[0] < m;
  equalM[0] = a[0] == m;
  moreM[0] = a[0] > m;
  for (long long i = (1); i < (n); i++) {
    lessM[i] = (a[i] < m) + lessM[i - 1];
    equalM[i] = (a[i] == m) + equalM[i - 1];
    moreM[i] = (a[i] > m) + moreM[i - 1];
  }
  long long penalty[n];
  for (long long i = (0); i < (n); i++) penalty[i] = moreM[i] - lessM[i];
  long long ans = 0;
  long long nearestM = -1;
  map<long long, long long> pos;
  pos[0] = 1;
  for (long long r = (0); r < (n); r++) {
    if (a[r] == m) {
      for (long long i = (nearestM + (nearestM == -1)); i < (r); i++)
        pos[penalty[i]]++;
      nearestM = r;
    }
    if (nearestM == -1) continue;
    ans += pos[penalty[r]] + pos[penalty[r] - 1];
  }
  cout << ans;
}
int32_t main() {
  srand(time(0));
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed;
  cout << setprecision(15);
  run();
  return 0;
}

```