#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, m, mIndex = -1;
  long long result = 0;
  cin >> n >> m;
  vector<int> numbers(n + 1);
  vector<int> lower(n + 1);
  vector<int> greater(n + 1);
  map<int, int> diffs;
  for (int it = 1; it <= n; ++it) {
    cin >> numbers[it];
    if (m == numbers[it]) {
      mIndex = it;
    }
  }
  diffs[0] = 1;
  for (int it = mIndex - 1; 0 < it; --it) {
    lower[it] = lower[it + 1];
    greater[it] = greater[it + 1];
    if (numbers[it] < m)
      ++lower[it];
    else
      ++greater[it];
    int diff = lower[it] - greater[it];
    diffs[diff] += 1;
  }
  for (int it = mIndex; it <= n; ++it) {
    lower[it] = lower[it - 1];
    greater[it] = greater[it - 1];
    greater[mIndex] = lower[mIndex] = 0;
    if (numbers[it] < m) ++lower[it];
    if (numbers[it] > m) ++greater[it];
    int diff = lower[it] - greater[it];
    result += diffs[-diff] + diffs[-diff - 1];
  }
  cout << result << endl;
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
int main() {
  int n, m, mIndex = -1;
  long long result = 0;
  cin >> n >> m;
  vector<int> numbers(n + 1);
  vector<int> lower(n + 1);
  vector<int> greater(n + 1);
  map<int, int> diffs;
  for (int it = 1; it <= n; ++it) {
    cin >> numbers[it];
    if (m == numbers[it]) {
      mIndex = it;
    }
  }
  diffs[0] = 1;
  for (int it = mIndex - 1; 0 < it; --it) {
    lower[it] = lower[it + 1];
    greater[it] = greater[it + 1];
    if (numbers[it] < m)
      ++lower[it];
    else
      ++greater[it];
    int diff = lower[it] - greater[it];
    diffs[diff] += 1;
  }
  for (int it = mIndex; it <= n; ++it) {
    lower[it] = lower[it - 1];
    greater[it] = greater[it - 1];
    greater[mIndex] = lower[mIndex] = 0;
    if (numbers[it] < m) ++lower[it];
    if (numbers[it] > m) ++greater[it];
    int diff = lower[it] - greater[it];
    result += diffs[-diff] + diffs[-diff - 1];
  }
  cout << result << endl;
  return 0;
}

```