#include <bits/stdc++.h>
using namespace std;
template <typename T>
T gcd(T x, T y) {
  if (y == 0)
    return x;
  else
    return gcd(y, x % y);
}
const double INF = 1e+9;
const double INFLL = 1e+18;
const double EPS = 1e-9;
const double PI = acos(-1);
const int knightDir[8][2] = {{-2, -1}, {-2, 1}, {-1, 2},  {1, 2},
                             {2, -1},  {2, 1},  {-1, -2}, {1, -2}};
const int dx[] = {-1, 0, 1, 0};
const int dy[] = {0, 1, 0, -1};
int l, m;
int pathMn;
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  int arr[n];
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i]);
  }
  int x = 0, sum = 0;
  bool flag = false;
  map<int, int> mp;
  mp[0] = 1;
  long long int ans = 0;
  for (int i = 0; i < n; i++) {
    if (arr[i] < k)
      sum--;
    else if (arr[i] > k)
      sum++;
    else
      flag = true;
    if (flag)
      ans += mp[sum] + mp[sum - 1];
    else
      mp[sum]++;
  }
  printf("%lld\n", ans);
}

### Prompt

Your task is to create 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;
template <typename T>
T gcd(T x, T y) {
  if (y == 0)
    return x;
  else
    return gcd(y, x % y);
}
const double INF = 1e+9;
const double INFLL = 1e+18;
const double EPS = 1e-9;
const double PI = acos(-1);
const int knightDir[8][2] = {{-2, -1}, {-2, 1}, {-1, 2},  {1, 2},
                             {2, -1},  {2, 1},  {-1, -2}, {1, -2}};
const int dx[] = {-1, 0, 1, 0};
const int dy[] = {0, 1, 0, -1};
int l, m;
int pathMn;
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  int arr[n];
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i]);
  }
  int x = 0, sum = 0;
  bool flag = false;
  map<int, int> mp;
  mp[0] = 1;
  long long int ans = 0;
  for (int i = 0; i < n; i++) {
    if (arr[i] < k)
      sum--;
    else if (arr[i] > k)
      sum++;
    else
      flag = true;
    if (flag)
      ans += mp[sum] + mp[sum - 1];
    else
      mp[sum]++;
  }
  printf("%lld\n", ans);
}

```