#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e9 + 7;
const int N = 4e5 + 10;
class BIT {
  int a[N];

 public:
  void init() { memset((a), (0), sizeof(a)); }
  inline int lowbit(int x) { return x & -x; }
  int sum(int x) {
    int ret = 0;
    while (x) {
      ret += a[x];
      x -= lowbit(x);
    }
    return ret;
  }
  void add(int x, int val) {
    while (x < N) {
      a[x] += val;
      x += lowbit(x);
    }
  }
} b;
int a[N];
int c[N];
long long getans(int n, int m) {
  int minv = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] <= m)
      c[i] = 1;
    else
      c[i] = -1;
    c[i] += c[i - 1];
    minv = min(minv, c[i]);
  }
  b.init();
  for (int i = 1; i <= n; i++) {
    c[i] -= minv - 1;
  }
  long long ans = 0;
  b.add(-minv + 1, 1);
  for (int i = 1; i <= n; i++) {
    b.add(c[i], 1);
    ans += b.sum(c[i]);
  }
  return ans;
}
int main() {
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i];
  cout << getans(n, m) - getans(n, m - 1) << endl;
  return 0;
}

### Prompt

In CPP, your task is to solve 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 long long INF = 1e9 + 7;
const int N = 4e5 + 10;
class BIT {
  int a[N];

 public:
  void init() { memset((a), (0), sizeof(a)); }
  inline int lowbit(int x) { return x & -x; }
  int sum(int x) {
    int ret = 0;
    while (x) {
      ret += a[x];
      x -= lowbit(x);
    }
    return ret;
  }
  void add(int x, int val) {
    while (x < N) {
      a[x] += val;
      x += lowbit(x);
    }
  }
} b;
int a[N];
int c[N];
long long getans(int n, int m) {
  int minv = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] <= m)
      c[i] = 1;
    else
      c[i] = -1;
    c[i] += c[i - 1];
    minv = min(minv, c[i]);
  }
  b.init();
  for (int i = 1; i <= n; i++) {
    c[i] -= minv - 1;
  }
  long long ans = 0;
  b.add(-minv + 1, 1);
  for (int i = 1; i <= n; i++) {
    b.add(c[i], 1);
    ans += b.sum(c[i]);
  }
  return ans;
}
int main() {
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i];
  cout << getans(n, m) - getans(n, m - 1) << endl;
  return 0;
}

```