#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 200 * 1000;
int p[MAX_N];
int n, m, poi;
long long ans;
void readInput() {
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    cin >> p[i];
    if (p[i] == m) poi = i;
  }
}
void update(vector<pair<int, int>> &vec) {
  int x = 0;
  for (int i = 1; i < (int)vec.size(); i++)
    if (vec[x].first == vec[i].first)
      vec[x].second++;
    else {
      vec[x + 1] = vec[i];
      x++;
    }
  while ((int)vec.size() - 1 > x) vec.pop_back();
}
void solve() {
  vector<pair<int, int>> left = {{0, 1}}, right = {{0, 1}};
  int tmp = 0;
  for (int i = poi - 1; i >= 0; i--) {
    if (p[i] > m)
      tmp++;
    else
      tmp--;
    left.push_back({tmp, 1});
  }
  tmp = 0;
  for (int i = poi + 1; i < n; i++) {
    if (p[i] > m)
      tmp++;
    else
      tmp--;
    right.push_back({tmp, 1});
  }
  sort(left.begin(), left.end());
  sort(right.begin(), right.end(), greater<pair<int, int>>());
  update(right);
  update(left);
  int cur = 0;
  for (auto i : left) {
    while (i.first + right[cur].first > 1 && cur < (int)right.size() - 1) cur++;
    if (i.first + right[cur].first == 1)
      ans += 1LL * right[cur].second * i.second, cur++;
    if (i.first + right[cur].first == 0)
      ans += 1LL * right[cur].second * i.second;
  }
}
int main() {
  readInput();
  solve();
  cout << ans;
}

### 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;
const int MAX_N = 200 * 1000;
int p[MAX_N];
int n, m, poi;
long long ans;
void readInput() {
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    cin >> p[i];
    if (p[i] == m) poi = i;
  }
}
void update(vector<pair<int, int>> &vec) {
  int x = 0;
  for (int i = 1; i < (int)vec.size(); i++)
    if (vec[x].first == vec[i].first)
      vec[x].second++;
    else {
      vec[x + 1] = vec[i];
      x++;
    }
  while ((int)vec.size() - 1 > x) vec.pop_back();
}
void solve() {
  vector<pair<int, int>> left = {{0, 1}}, right = {{0, 1}};
  int tmp = 0;
  for (int i = poi - 1; i >= 0; i--) {
    if (p[i] > m)
      tmp++;
    else
      tmp--;
    left.push_back({tmp, 1});
  }
  tmp = 0;
  for (int i = poi + 1; i < n; i++) {
    if (p[i] > m)
      tmp++;
    else
      tmp--;
    right.push_back({tmp, 1});
  }
  sort(left.begin(), left.end());
  sort(right.begin(), right.end(), greater<pair<int, int>>());
  update(right);
  update(left);
  int cur = 0;
  for (auto i : left) {
    while (i.first + right[cur].first > 1 && cur < (int)right.size() - 1) cur++;
    if (i.first + right[cur].first == 1)
      ans += 1LL * right[cur].second * i.second, cur++;
    if (i.first + right[cur].first == 0)
      ans += 1LL * right[cur].second * i.second;
  }
}
int main() {
  readInput();
  solve();
  cout << ans;
}

```