#include <bits/stdc++.h>
using namespace std;
char buf[25];
const int maxn = 200010;
int a[maxn], cnt[maxn * 2];
int n, m;
int read() {
  int x = 0, f = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = 1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return f ? -x : x;
}
void write(long long x) {
  if (!x) {
    putchar('0');
    return;
  }
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  int cnt = 0;
  while (x) {
    buf[++cnt] = '0' + x % 10;
    x /= 10;
  }
  for (int i = cnt; i >= 1; --i) putchar(buf[i]);
}
long long solve(int x) {
  for (int i = 0; i <= n * 2; ++i) cnt[i] = 0;
  cnt[n] = 1;
  int s = n;
  long long sum = 0, ans = 0;
  if (a[1] <= x) {
    ++sum;
    ++ans;
    ++cnt[++s];
  } else
    ++cnt[--s];
  for (int i = 2; i <= n; ++i) {
    if (a[i] <= x)
      sum += cnt[++s];
    else
      sum -= cnt[s--];
    ans += sum;
    ++cnt[s];
  }
  return ans;
}
int main() {
  n = read();
  m = read();
  for (int i = 1; i <= n; ++i) a[i] = read();
  write(solve(m) - solve(m - 1));
  putchar('\n');
  return 0;
}

### Prompt

Your challenge is to write 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;
char buf[25];
const int maxn = 200010;
int a[maxn], cnt[maxn * 2];
int n, m;
int read() {
  int x = 0, f = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = 1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return f ? -x : x;
}
void write(long long x) {
  if (!x) {
    putchar('0');
    return;
  }
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  int cnt = 0;
  while (x) {
    buf[++cnt] = '0' + x % 10;
    x /= 10;
  }
  for (int i = cnt; i >= 1; --i) putchar(buf[i]);
}
long long solve(int x) {
  for (int i = 0; i <= n * 2; ++i) cnt[i] = 0;
  cnt[n] = 1;
  int s = n;
  long long sum = 0, ans = 0;
  if (a[1] <= x) {
    ++sum;
    ++ans;
    ++cnt[++s];
  } else
    ++cnt[--s];
  for (int i = 2; i <= n; ++i) {
    if (a[i] <= x)
      sum += cnt[++s];
    else
      sum -= cnt[s--];
    ans += sum;
    ++cnt[s];
  }
  return ans;
}
int main() {
  n = read();
  m = read();
  for (int i = 1; i <= n; ++i) a[i] = read();
  write(solve(m) - solve(m - 1));
  putchar('\n');
  return 0;
}

```