#include <bits/stdc++.h>
using namespace std;
const long long int mod = 1e9 + 9;
bool IsPowerOfTwo(long long int x) { return (x != 0) && ((x & (x - 1)) == 0); }
long long int poer(long long int a, long long int b) {
  long long int rs = 1;
  while (b > 0) {
    if (b & 1) rs *= a;
    a *= a;
    b /= 2;
  }
  return rs;
}
long long int mod_poer(long long int a, long long int b, long long int mo) {
  long long int rs = 1;
  while (b > 0) {
    if (b & 1) rs = (rs * a) % mo;
    a = (a * a) % mo;
    b /= 2;
  }
  return rs % mo;
}
long long int d, x, y;
void ext_euclid(long long int a, long long int b) {
  if (b == 0) {
    d = a;
    x = 1;
    y = 0;
  } else {
    ext_euclid(b, a % b);
    long long int temp = x;
    x = y;
    y = temp - (a / b) * y;
  }
}
long long int mod_inverse1(long long int a, long long int m) {
  ext_euclid(a, m);
  return (x % m + m) % m;
}
long long int mod_inverse2(long long int a, long long int m) {
  return mod_poer(a, m - 2, m);
}
long long int gcd(long long int a, long long int b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long int fact[200005];
void fac() {
  fact[1] = fact[0] = 1;
  for (long long int i = 2; i <= 200005; ++i) {
    fact[i] = ((fact[i - 1] % mod) * i) % mod;
  }
}
long long int combi(long long int a, long long int b) {
  if (b > a) return 0;
  if (b == a)
    return 1;
  else {
    long long int x = mod_poer(fact[b], mod - 2, mod) % mod;
    long long int y = mod_poer(fact[a - b], mod - 2, mod) % mod;
    return ((((fact[a]) * x) % mod) * y) % mod;
  }
}
long long int arr[200005], n, m;
map<long long int, long long int> vis;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  for (long long int i = 1; i <= n; ++i) cin >> arr[i];
  long long int pos = 0;
  for (long long int i = 1; i <= n; ++i)
    if (arr[i] == m) {
      pos = i;
      break;
    }
  long long int cnt = 0;
  for (long long int i = pos; i <= n; ++i) {
    if (arr[i] > m) cnt++;
    if (arr[i] < m) cnt--;
    vis[cnt]++;
  }
  cnt = 0;
  long long int ans = 0;
  for (long long int i = pos; i >= 1; i--) {
    if (arr[i] > m) cnt--;
    if (arr[i] < m) cnt++;
    ans += (vis[cnt] + vis[cnt + 1]);
  }
  cout << ans << "\n";
  return 0;
}

### Prompt

Generate 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;
const long long int mod = 1e9 + 9;
bool IsPowerOfTwo(long long int x) { return (x != 0) && ((x & (x - 1)) == 0); }
long long int poer(long long int a, long long int b) {
  long long int rs = 1;
  while (b > 0) {
    if (b & 1) rs *= a;
    a *= a;
    b /= 2;
  }
  return rs;
}
long long int mod_poer(long long int a, long long int b, long long int mo) {
  long long int rs = 1;
  while (b > 0) {
    if (b & 1) rs = (rs * a) % mo;
    a = (a * a) % mo;
    b /= 2;
  }
  return rs % mo;
}
long long int d, x, y;
void ext_euclid(long long int a, long long int b) {
  if (b == 0) {
    d = a;
    x = 1;
    y = 0;
  } else {
    ext_euclid(b, a % b);
    long long int temp = x;
    x = y;
    y = temp - (a / b) * y;
  }
}
long long int mod_inverse1(long long int a, long long int m) {
  ext_euclid(a, m);
  return (x % m + m) % m;
}
long long int mod_inverse2(long long int a, long long int m) {
  return mod_poer(a, m - 2, m);
}
long long int gcd(long long int a, long long int b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long int fact[200005];
void fac() {
  fact[1] = fact[0] = 1;
  for (long long int i = 2; i <= 200005; ++i) {
    fact[i] = ((fact[i - 1] % mod) * i) % mod;
  }
}
long long int combi(long long int a, long long int b) {
  if (b > a) return 0;
  if (b == a)
    return 1;
  else {
    long long int x = mod_poer(fact[b], mod - 2, mod) % mod;
    long long int y = mod_poer(fact[a - b], mod - 2, mod) % mod;
    return ((((fact[a]) * x) % mod) * y) % mod;
  }
}
long long int arr[200005], n, m;
map<long long int, long long int> vis;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  for (long long int i = 1; i <= n; ++i) cin >> arr[i];
  long long int pos = 0;
  for (long long int i = 1; i <= n; ++i)
    if (arr[i] == m) {
      pos = i;
      break;
    }
  long long int cnt = 0;
  for (long long int i = pos; i <= n; ++i) {
    if (arr[i] > m) cnt++;
    if (arr[i] < m) cnt--;
    vis[cnt]++;
  }
  cnt = 0;
  long long int ans = 0;
  for (long long int i = pos; i >= 1; i--) {
    if (arr[i] > m) cnt--;
    if (arr[i] < m) cnt++;
    ans += (vis[cnt] + vis[cnt + 1]);
  }
  cout << ans << "\n";
  return 0;
}

```