#include <bits/stdc++.h>
using namespace std;
template <typename T, typename U>
inline bool smax(T& x, const U& y) {
  return y > x ? (x = y, 1) : 0;
}
template <typename T, typename U>
inline bool smin(T& x, const U& y) {
  return y < x ? (x = y, 1) : 0;
}
inline int read() {
  int w = 0, p = 0;
  char c;
  while (!isdigit(c = getchar()))
    if (c == '-') p = 1;
  while (isdigit(c)) w = (w << 1) + (w << 3) + (c ^ '0'), c = getchar();
  return p ? -w : w;
}
const int N = 4e5 + 5;
int n, a[N], c[N];
;
inline void add(int p) {
  p += n + 1;
  while (p <= n + n) ++c[p], p += p & -p;
}
inline int ask(int p) {
  p += n + 1;
  int ans = 0;
  while (p) ans += c[p], p -= p & -p;
  return ans;
}
inline long long solve(int x) {
  long long ans = 0;
  int sum = 0;
  memset(c, 0, sizeof c);
  add(0);
  for (int i = (1); i <= (n); ++i) {
    if (a[i] > x)
      ++sum;
    else
      --sum;
    ans += ask(-sum);
    add(-sum);
  }
  return ans;
}
int main() {
  n = read();
  int m = read();
  for (int i = (1); i <= (n); ++i) a[i] = read();
  cout << solve(m) - solve(m - 1);
  return 0;
}

### Prompt

Create a solution in cpp for 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, typename U>
inline bool smax(T& x, const U& y) {
  return y > x ? (x = y, 1) : 0;
}
template <typename T, typename U>
inline bool smin(T& x, const U& y) {
  return y < x ? (x = y, 1) : 0;
}
inline int read() {
  int w = 0, p = 0;
  char c;
  while (!isdigit(c = getchar()))
    if (c == '-') p = 1;
  while (isdigit(c)) w = (w << 1) + (w << 3) + (c ^ '0'), c = getchar();
  return p ? -w : w;
}
const int N = 4e5 + 5;
int n, a[N], c[N];
;
inline void add(int p) {
  p += n + 1;
  while (p <= n + n) ++c[p], p += p & -p;
}
inline int ask(int p) {
  p += n + 1;
  int ans = 0;
  while (p) ans += c[p], p -= p & -p;
  return ans;
}
inline long long solve(int x) {
  long long ans = 0;
  int sum = 0;
  memset(c, 0, sizeof c);
  add(0);
  for (int i = (1); i <= (n); ++i) {
    if (a[i] > x)
      ++sum;
    else
      --sum;
    ans += ask(-sum);
    add(-sum);
  }
  return ans;
}
int main() {
  n = read();
  int m = read();
  for (int i = (1); i <= (n); ++i) a[i] = read();
  cout << solve(m) - solve(m - 1);
  return 0;
}

```