#include <bits/stdc++.h>
using namespace std;
long long int gcd(long long int a, long long int b) {
  if (b < a) {
    long long int temp = a;
    a = b;
    b = temp;
  }
  if (a == 0) {
    return b;
  }
  return gcd(b % a, a);
}
long long int max(long long int a, long long int b) {
  if (a > b) return a;
  return b;
}
long long int lcm(long long int a, long long int b) {
  return (a * b / gcd(a, b));
}
long long int fact(long long int x) {
  long long int ans = 1;
  if (x == 0) {
    return 1;
  }
  while (x > 0) {
    ans = (ans * x) % 1000000007;
    x--;
  }
  return ans % 1000000007;
}
long long int comb(long long int n, long long int r) {
  long long int x = fact(r), y = fact(n - r);
  x = (x * y) % 1000000007;
  if (fact(n) > x) return fact(n) / x;
  return (fact(n) + 1000000007) / x;
}
long long int power(long long int a, long long int p) {
  long long int ans = 1;
  while (p > 0) {
    if (p % 2) {
      ans *= a;
      ans %= 1000000007;
    }
    a *= a;
    a %= 1000000007;
    p /= 2;
  }
  return ans % 1000000007;
}
long long int dist(long long int x1, long long int y1, long long int x2,
                   long long int y2) {
  return (1.0 * (x1 - x2) * (x1 - x2) + 1.0 * (y1 - y2) * (y1 - y2));
}
vector<long long int> getbinary(long long int x) {
  vector<long long int> bin(20000, 0);
  long long int iter = 0;
  while (x > 0) {
    if (x % 2 == 1) {
      bin[iter] = 1;
    } else {
      bin[iter] = 0;
    }
    iter++;
    x /= 2;
  }
  return bin;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, m, i, x, g = 0, s = 0, t = 0;
  cin >> n >> m;
  long long int a[n];
  map<long long int, long long int> b;
  for (i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] == m) {
      x = i;
    }
  }
  b[0] = 1;
  for (i = x - 1; i >= 0; i--) {
    if (a[i] > m)
      g++;
    else
      s++;
    b[g - s]++;
  }
  g = s = 0;
  for (i = x; i < n; i++) {
    if (a[i] > m)
      g++;
    else if (a[i] < m)
      s++;
    t += b[s - g];
    t += b[s - g + 1];
  }
  cout << t << endl;
}

### 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;
long long int gcd(long long int a, long long int b) {
  if (b < a) {
    long long int temp = a;
    a = b;
    b = temp;
  }
  if (a == 0) {
    return b;
  }
  return gcd(b % a, a);
}
long long int max(long long int a, long long int b) {
  if (a > b) return a;
  return b;
}
long long int lcm(long long int a, long long int b) {
  return (a * b / gcd(a, b));
}
long long int fact(long long int x) {
  long long int ans = 1;
  if (x == 0) {
    return 1;
  }
  while (x > 0) {
    ans = (ans * x) % 1000000007;
    x--;
  }
  return ans % 1000000007;
}
long long int comb(long long int n, long long int r) {
  long long int x = fact(r), y = fact(n - r);
  x = (x * y) % 1000000007;
  if (fact(n) > x) return fact(n) / x;
  return (fact(n) + 1000000007) / x;
}
long long int power(long long int a, long long int p) {
  long long int ans = 1;
  while (p > 0) {
    if (p % 2) {
      ans *= a;
      ans %= 1000000007;
    }
    a *= a;
    a %= 1000000007;
    p /= 2;
  }
  return ans % 1000000007;
}
long long int dist(long long int x1, long long int y1, long long int x2,
                   long long int y2) {
  return (1.0 * (x1 - x2) * (x1 - x2) + 1.0 * (y1 - y2) * (y1 - y2));
}
vector<long long int> getbinary(long long int x) {
  vector<long long int> bin(20000, 0);
  long long int iter = 0;
  while (x > 0) {
    if (x % 2 == 1) {
      bin[iter] = 1;
    } else {
      bin[iter] = 0;
    }
    iter++;
    x /= 2;
  }
  return bin;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, m, i, x, g = 0, s = 0, t = 0;
  cin >> n >> m;
  long long int a[n];
  map<long long int, long long int> b;
  for (i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] == m) {
      x = i;
    }
  }
  b[0] = 1;
  for (i = x - 1; i >= 0; i--) {
    if (a[i] > m)
      g++;
    else
      s++;
    b[g - s]++;
  }
  g = s = 0;
  for (i = x; i < n; i++) {
    if (a[i] > m)
      g++;
    else if (a[i] < m)
      s++;
    t += b[s - g];
    t += b[s - g + 1];
  }
  cout << t << endl;
}

```