#include <bits/stdc++.h>
using namespace std;
vector<int> factorize(int n) {
  vector<int> res;
  for (int i = 2; i * i <= n; ++i) {
    while (n % i == 0) {
      res.push_back(i);
      n /= i;
    }
  }
  if (n != 1) {
    res.push_back(n);
  }
  return res;
}
long long int multiple(long long int a, long long int b, long long int c) {
  if (b == 0) return 0;
  long long int ret = multiple(a, b >> 1, c);
  ret = (ret + ret) % c;
  if (b & 1) ret = (ret + a) % c;
  return ret;
}
long int GCD(int A, int B) {
  if (B == 0)
    return A;
  else
    return GCD(B, A % B);
}
long int Log2(long int x) {
  long int ans = 0;
  while (x >>= 1) ans++;
  return ans;
}
long long int mini(long long int a, long long int b) {
  if (a <= b)
    return a;
  else
    return b;
}
long long int maxe(long long int a, long long int b) {
  if (a >= b)
    return a;
  else
    return b;
}
int main() {
  long long int n, m;
  cin >> n >> m;
  set<long long int> s;
  vector<long long int> v(n + 1);
  long long int a, b, c, d, ind;
  for (long long int i = 1; i <= n; i++) {
    cin >> v[i];
    if (v[i] == m) ind = i;
  }
  long long int chote = 0, bade = 0, ans = 0;
  map<long long int, long long int> mr;
  for (long int i = ind - 1; i >= 1; i--) {
    if (v[i] < m)
      chote++;
    else
      bade++;
    if (chote == bade - 1)
      ans++;
    else if (chote == bade)
      ans++;
    s.insert(chote - bade);
    mr[chote - bade]++;
  }
  chote = 0, bade = 0;
  for (long int i = ind + 1; i <= n; i++) {
    if (v[i] < m)
      chote++;
    else
      bade++;
    if (chote == bade - 1)
      ans++;
    else if (chote == bade)
      ans++;
    if (s.find(bade - chote) != s.end()) ans += mr[bade - chote];
    if (s.find(bade - 1 - chote) != s.end()) ans += mr[bade - 1 - chote];
  }
  cout << ans + 1;
}

### 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;
vector<int> factorize(int n) {
  vector<int> res;
  for (int i = 2; i * i <= n; ++i) {
    while (n % i == 0) {
      res.push_back(i);
      n /= i;
    }
  }
  if (n != 1) {
    res.push_back(n);
  }
  return res;
}
long long int multiple(long long int a, long long int b, long long int c) {
  if (b == 0) return 0;
  long long int ret = multiple(a, b >> 1, c);
  ret = (ret + ret) % c;
  if (b & 1) ret = (ret + a) % c;
  return ret;
}
long int GCD(int A, int B) {
  if (B == 0)
    return A;
  else
    return GCD(B, A % B);
}
long int Log2(long int x) {
  long int ans = 0;
  while (x >>= 1) ans++;
  return ans;
}
long long int mini(long long int a, long long int b) {
  if (a <= b)
    return a;
  else
    return b;
}
long long int maxe(long long int a, long long int b) {
  if (a >= b)
    return a;
  else
    return b;
}
int main() {
  long long int n, m;
  cin >> n >> m;
  set<long long int> s;
  vector<long long int> v(n + 1);
  long long int a, b, c, d, ind;
  for (long long int i = 1; i <= n; i++) {
    cin >> v[i];
    if (v[i] == m) ind = i;
  }
  long long int chote = 0, bade = 0, ans = 0;
  map<long long int, long long int> mr;
  for (long int i = ind - 1; i >= 1; i--) {
    if (v[i] < m)
      chote++;
    else
      bade++;
    if (chote == bade - 1)
      ans++;
    else if (chote == bade)
      ans++;
    s.insert(chote - bade);
    mr[chote - bade]++;
  }
  chote = 0, bade = 0;
  for (long int i = ind + 1; i <= n; i++) {
    if (v[i] < m)
      chote++;
    else
      bade++;
    if (chote == bade - 1)
      ans++;
    else if (chote == bade)
      ans++;
    if (s.find(bade - chote) != s.end()) ans += mr[bade - chote];
    if (s.find(bade - 1 - chote) != s.end()) ans += mr[bade - 1 - chote];
  }
  cout << ans + 1;
}

```