#include <bits/stdc++.h>
using namespace std;
long long read() {
  long long w = 0, f = 1;
  char c = ' ';
  while (c < '0' || c > '9') c = getchar(), f = c == '-' ? -1 : f;
  while (c >= '0' && c <= '9') w = w * 10 + c - 48, c = getchar();
  return w * f;
}
long long val[200000 << 3 | 5];
void update(long long p, long long x, long long v, long long l, long long r) {
  if (l == r) {
    val[p] += v;
    return;
  }
  long long mid = (l + r) >> 1;
  if (x <= mid)
    update(p << 1, x, v, l, mid);
  else
    update(p << 1 | 1, x, v, mid + 1, r);
  val[p] = val[p << 1] + val[p << 1 | 1];
}
long long query(long long p, long long x, long long y, long long l,
                long long r) {
  if (x <= l && y >= r) return val[p];
  long long mid = (l + r) >> 1;
  if (y <= mid) return query(p << 1, x, y, l, mid);
  if (x > mid) return query(p << 1 | 1, x, y, mid + 1, r);
  return query(p << 1, x, y, l, mid) + query(p << 1 | 1, x, y, mid + 1, r);
}
long long n, m, a[200005];
long long solve(long long x) {
  memset(val, 0, sizeof(val));
  long long sum = 0, res = 0;
  for (long long i = 1; i <= n; i++) {
    update(1, 200000 - sum + 1, 1, 1, 200000 << 1);
    sum += (a[i] > x ? 1 : -1);
    res += query(1, 1, 200000 - sum + 1, 1, 200000 << 1);
  }
  return res;
}
signed main() {
  n = read(), m = read();
  for (long long i = 1; i <= n; i++) a[i] = read();
  printf("%lld", solve(m) - solve(m - 1));
  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;
long long read() {
  long long w = 0, f = 1;
  char c = ' ';
  while (c < '0' || c > '9') c = getchar(), f = c == '-' ? -1 : f;
  while (c >= '0' && c <= '9') w = w * 10 + c - 48, c = getchar();
  return w * f;
}
long long val[200000 << 3 | 5];
void update(long long p, long long x, long long v, long long l, long long r) {
  if (l == r) {
    val[p] += v;
    return;
  }
  long long mid = (l + r) >> 1;
  if (x <= mid)
    update(p << 1, x, v, l, mid);
  else
    update(p << 1 | 1, x, v, mid + 1, r);
  val[p] = val[p << 1] + val[p << 1 | 1];
}
long long query(long long p, long long x, long long y, long long l,
                long long r) {
  if (x <= l && y >= r) return val[p];
  long long mid = (l + r) >> 1;
  if (y <= mid) return query(p << 1, x, y, l, mid);
  if (x > mid) return query(p << 1 | 1, x, y, mid + 1, r);
  return query(p << 1, x, y, l, mid) + query(p << 1 | 1, x, y, mid + 1, r);
}
long long n, m, a[200005];
long long solve(long long x) {
  memset(val, 0, sizeof(val));
  long long sum = 0, res = 0;
  for (long long i = 1; i <= n; i++) {
    update(1, 200000 - sum + 1, 1, 1, 200000 << 1);
    sum += (a[i] > x ? 1 : -1);
    res += query(1, 1, 200000 - sum + 1, 1, 200000 << 1);
  }
  return res;
}
signed main() {
  n = read(), m = read();
  for (long long i = 1; i <= n; i++) a[i] = read();
  printf("%lld", solve(m) - solve(m - 1));
  return 0;
}

```