#include <bits/stdc++.h>
using namespace std;
bool OUT = false;
template <class T>
void out_vector(vector<T> v) {
  cout << "------------------" << endl;
  for (int j = 0; j < v.size(); j++) {
    cout << v[j] << " ";
  }
  cout << endl << "------------------" << endl;
}
int main() {
  long long n, m, i, j, m_index, count = 1, bigger, smaller, diff;
  cin >> n >> m;
  vector<long long> v(n);
  for (i = 0; i < n; i++) {
    cin >> v[i];
    if (v[i] == m) m_index = i;
  }
  vector<long long> left, right;
  map<long long, long long> m_left, m_right;
  bigger = 0, smaller = 0;
  for (i = m_index - 1; i >= 0; i--) {
    if (v[i] > m) bigger++;
    if (v[i] < m) smaller++;
    if (bigger == smaller || bigger == smaller + 1) {
      if (OUT) {
        cout << "for left " << i << ", bigger " << bigger << ", smaller "
             << smaller << ", count: " << count << endl;
      }
      count++;
    }
    left.push_back(bigger - smaller);
    m_left[bigger - smaller]++;
  }
  bigger = 0, smaller = 0;
  for (i = m_index + 1; i < n; i++) {
    if (v[i] > m) bigger++;
    if (v[i] < m) smaller++;
    if (bigger == smaller || bigger == smaller + 1) {
      if (OUT) {
        cout << "for right " << i << ", bigger " << bigger << ", smaller "
             << smaller << ", count: " << count << endl;
      }
      count++;
    }
    right.push_back(bigger - smaller);
    m_right[bigger - smaller]++;
  }
  long long sum;
  for (auto &diff : m_left) {
    sum = m_right[0 - diff.first] * diff.second;
    count += sum;
    sum = m_right[1 - diff.first] * diff.second;
    count += sum;
  }
  if (OUT) out_vector(left);
  if (OUT) out_vector(right);
  cout << count;
  return 0;
}

### Prompt

Your challenge is to write 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;
bool OUT = false;
template <class T>
void out_vector(vector<T> v) {
  cout << "------------------" << endl;
  for (int j = 0; j < v.size(); j++) {
    cout << v[j] << " ";
  }
  cout << endl << "------------------" << endl;
}
int main() {
  long long n, m, i, j, m_index, count = 1, bigger, smaller, diff;
  cin >> n >> m;
  vector<long long> v(n);
  for (i = 0; i < n; i++) {
    cin >> v[i];
    if (v[i] == m) m_index = i;
  }
  vector<long long> left, right;
  map<long long, long long> m_left, m_right;
  bigger = 0, smaller = 0;
  for (i = m_index - 1; i >= 0; i--) {
    if (v[i] > m) bigger++;
    if (v[i] < m) smaller++;
    if (bigger == smaller || bigger == smaller + 1) {
      if (OUT) {
        cout << "for left " << i << ", bigger " << bigger << ", smaller "
             << smaller << ", count: " << count << endl;
      }
      count++;
    }
    left.push_back(bigger - smaller);
    m_left[bigger - smaller]++;
  }
  bigger = 0, smaller = 0;
  for (i = m_index + 1; i < n; i++) {
    if (v[i] > m) bigger++;
    if (v[i] < m) smaller++;
    if (bigger == smaller || bigger == smaller + 1) {
      if (OUT) {
        cout << "for right " << i << ", bigger " << bigger << ", smaller "
             << smaller << ", count: " << count << endl;
      }
      count++;
    }
    right.push_back(bigger - smaller);
    m_right[bigger - smaller]++;
  }
  long long sum;
  for (auto &diff : m_left) {
    sum = m_right[0 - diff.first] * diff.second;
    count += sum;
    sum = m_right[1 - diff.first] * diff.second;
    count += sum;
  }
  if (OUT) out_vector(left);
  if (OUT) out_vector(right);
  cout << count;
  return 0;
}

```