#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + c - '0';
    c = getchar();
  }
  return x * f;
}
int n, m, a[200010];
struct BIT {
  int b[200010 << 1];
  void clear() { memset(b, 0, sizeof(b)); }
  inline int lowbit(int x) { return x & (-x); }
  inline void Add(int x) {
    while (x <= (n << 1 | 1)) ++b[x], x += lowbit(x);
  }
  inline int Ask(int x) {
    int ans = 0;
    while (x) ans += b[x], x -= lowbit(x);
    return ans;
  }
} B;
long long Solve(long long x) {
  B.clear();
  B.Add(n + 1);
  int now = 0;
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    now += a[i] >= x ? 1 : -1;
    ans += B.Ask(now + n - 1 + 1);
    B.Add(now + n + 1);
  }
  return ans;
}
int main() {
  n = read(), m = read();
  for (int i = 1; i <= n; ++i) {
    a[i] = read();
  }
  printf("%lld\n", Solve(m) - Solve(m + 1));
  return 0;
}

### Prompt

Generate 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;
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + c - '0';
    c = getchar();
  }
  return x * f;
}
int n, m, a[200010];
struct BIT {
  int b[200010 << 1];
  void clear() { memset(b, 0, sizeof(b)); }
  inline int lowbit(int x) { return x & (-x); }
  inline void Add(int x) {
    while (x <= (n << 1 | 1)) ++b[x], x += lowbit(x);
  }
  inline int Ask(int x) {
    int ans = 0;
    while (x) ans += b[x], x -= lowbit(x);
    return ans;
  }
} B;
long long Solve(long long x) {
  B.clear();
  B.Add(n + 1);
  int now = 0;
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    now += a[i] >= x ? 1 : -1;
    ans += B.Ask(now + n - 1 + 1);
    B.Add(now + n + 1);
  }
  return ans;
}
int main() {
  n = read(), m = read();
  for (int i = 1; i <= n; ++i) {
    a[i] = read();
  }
  printf("%lld\n", Solve(m) - Solve(m + 1));
  return 0;
}

```