#include <bits/stdc++.h>
using namespace std;
long long const mod = 1e9 + 7;
long long const inf = 2e18;
inline long long add(long long a, long long b) {
  return (a % mod + b % mod + mod) % mod;
}
inline long long mul(long long a, long long b) {
  return (a % mod * b % mod + mod) % mod;
}
inline long long sub(long long a, long long b) {
  return (a % mod - b % mod + mod) % mod;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long modexp(long long x, long long n) {
  if (n == 0) {
    return 1;
  } else if (n & 1) {
    return (modexp((x * x) % mod, n / 2) % mod * x % mod) % mod;
  } else {
    return (modexp((x * x) % mod, n / 2) % mod);
  }
}
long long modinv(long long n) { return modexp(n, mod - 2); }
void solve() {
  long long n, m;
  cin >> n >> m;
  long long a[n];
  for (long long i = 0; i < (n); i++) cin >> a[i];
  map<long long, long long> m1;
  long long r = 0, second = 0;
  int id = -1;
  for (int i = n - 1; i >= 0; i--) {
    if (a[i] == m) {
      id = i;
      break;
    }
  }
  for (int i = id + 1; i < n; i++) {
    if (a[i] < m)
      r++;
    else
      second++;
    if (a[i] == m) {
      break;
    }
    m1[second - r]++;
  }
  long long ans = 0;
  long long p = 0, q = 0;
  for (int i = id; i >= 0; i--) {
    if (a[i] < m)
      p++;
    else if (a[i] > m)
      q++;
    if (p == q || q == (p + 1)) ans++;
    ans = ans + (m1[p - q]);
    ans = ans + (m1[p - q + 1]);
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long T = 1;
  for (int t = 1; t <= T; t++) {
    solve();
  }
  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;
long long const mod = 1e9 + 7;
long long const inf = 2e18;
inline long long add(long long a, long long b) {
  return (a % mod + b % mod + mod) % mod;
}
inline long long mul(long long a, long long b) {
  return (a % mod * b % mod + mod) % mod;
}
inline long long sub(long long a, long long b) {
  return (a % mod - b % mod + mod) % mod;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long modexp(long long x, long long n) {
  if (n == 0) {
    return 1;
  } else if (n & 1) {
    return (modexp((x * x) % mod, n / 2) % mod * x % mod) % mod;
  } else {
    return (modexp((x * x) % mod, n / 2) % mod);
  }
}
long long modinv(long long n) { return modexp(n, mod - 2); }
void solve() {
  long long n, m;
  cin >> n >> m;
  long long a[n];
  for (long long i = 0; i < (n); i++) cin >> a[i];
  map<long long, long long> m1;
  long long r = 0, second = 0;
  int id = -1;
  for (int i = n - 1; i >= 0; i--) {
    if (a[i] == m) {
      id = i;
      break;
    }
  }
  for (int i = id + 1; i < n; i++) {
    if (a[i] < m)
      r++;
    else
      second++;
    if (a[i] == m) {
      break;
    }
    m1[second - r]++;
  }
  long long ans = 0;
  long long p = 0, q = 0;
  for (int i = id; i >= 0; i--) {
    if (a[i] < m)
      p++;
    else if (a[i] > m)
      q++;
    if (p == q || q == (p + 1)) ans++;
    ans = ans + (m1[p - q]);
    ans = ans + (m1[p - q + 1]);
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long T = 1;
  for (int t = 1; t <= T; t++) {
    solve();
  }
  return 0;
}

```