#include <bits/stdc++.h>
using namespace std;
const int maxn = 2 * 1e5 + 17;
long long ans;
int N, M;
int arr[maxn], where[maxn];
int l[2][maxn], r[2][maxn];
int net_left[maxn], net_right[maxn];
map<int, int> m;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> N >> M;
  for (int i = 1; i <= N; i++) cin >> arr[i], where[arr[i]] = i;
  int mid = where[M];
  for (int i = mid - 1; i; i--)
    l[0][i] = l[0][i + 1] + (arr[i] < M), l[1][i] = l[1][i + 1] + (arr[i] > M);
  for (int i = mid + 1; i <= N; i++)
    r[0][i] = r[0][i - 1] + (arr[i] < M), r[1][i] = r[1][i - 1] + (arr[i] > M);
  for (int i = 1; i < mid; i++) net_left[i] = l[1][i] - l[0][i];
  for (int i = mid + 1; i <= N; i++) net_right[i] = r[1][i] - r[0][i];
  for (int i = 1; i < mid; i++) ans += net_left[i] == 0 || net_left[i] == 1;
  for (int i = mid + 1; i <= N; i++)
    ans += net_right[i] == 0 || net_right[i] == 1;
  for (int i = mid + 1; i <= N; i++) m[net_right[i]]++;
  for (int i = mid - 1; i; i--) ans += m[-net_left[i]] + m[-net_left[i] + 1];
  cout << ++ans << 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;
const int maxn = 2 * 1e5 + 17;
long long ans;
int N, M;
int arr[maxn], where[maxn];
int l[2][maxn], r[2][maxn];
int net_left[maxn], net_right[maxn];
map<int, int> m;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> N >> M;
  for (int i = 1; i <= N; i++) cin >> arr[i], where[arr[i]] = i;
  int mid = where[M];
  for (int i = mid - 1; i; i--)
    l[0][i] = l[0][i + 1] + (arr[i] < M), l[1][i] = l[1][i + 1] + (arr[i] > M);
  for (int i = mid + 1; i <= N; i++)
    r[0][i] = r[0][i - 1] + (arr[i] < M), r[1][i] = r[1][i - 1] + (arr[i] > M);
  for (int i = 1; i < mid; i++) net_left[i] = l[1][i] - l[0][i];
  for (int i = mid + 1; i <= N; i++) net_right[i] = r[1][i] - r[0][i];
  for (int i = 1; i < mid; i++) ans += net_left[i] == 0 || net_left[i] == 1;
  for (int i = mid + 1; i <= N; i++)
    ans += net_right[i] == 0 || net_right[i] == 1;
  for (int i = mid + 1; i <= N; i++) m[net_right[i]]++;
  for (int i = mid - 1; i; i--) ans += m[-net_left[i]] + m[-net_left[i] + 1];
  cout << ++ans << endl;
  return 0;
}

```