#include <bits/stdc++.h>
inline long long MAX2(long long a, long long int b) {
  return (a) > (b) ? (a) : (b);
}
inline long long MAX3(long long a, long long b, long long c) {
  return (a) > (b) ? ((a) > (c) ? (a) : (c)) : ((b) > (c) ? (b) : (c));
}
inline long long MIN2(long long a, long long b) {
  return (a) < (b) ? (a) : (b);
}
inline long long MIN3(long long a, long long b, long long c) {
  return (a) < (b) ? ((a) < (c) ? (a) : (c)) : ((b) < (c) ? (b) : (c));
}
using namespace std;
int main() {
  int i;
  long long n, m;
  long long p[200005];
  cin >> n >> m;
  map<long long, long long> left, right;
  map<long long, long long>::iterator it, it1;
  long long mpos;
  for (i = 0; i < n; i++) {
    cin >> p[i];
    if (p[i] == m) mpos = i;
  }
  long long less = 0, more = 0;
  for (i = mpos - 1; i >= 0; i--) {
    if (p[i] < m)
      less++;
    else
      more++;
    left[less - more]++;
  }
  less = 0, more = 0;
  for (i = mpos + 1; i < n; i++) {
    if (p[i] < m)
      less++;
    else
      more++;
    right[less - more]++;
  }
  long long ans = 1;
  for (i = 0; i >= -1; i--) {
    if (left.find(i) != left.end()) ans += left[i];
    if (right.find(i) != right.end()) ans += right[i];
  }
  for (it = right.begin(); it != right.end(); it++) {
    for (i = 0; i >= -1; i--) {
      it1 = left.find(-it->first + i);
      if (it1 != left.end()) ans += it->second * it1->second;
    }
  }
  cout << ans;
}

### 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>
inline long long MAX2(long long a, long long int b) {
  return (a) > (b) ? (a) : (b);
}
inline long long MAX3(long long a, long long b, long long c) {
  return (a) > (b) ? ((a) > (c) ? (a) : (c)) : ((b) > (c) ? (b) : (c));
}
inline long long MIN2(long long a, long long b) {
  return (a) < (b) ? (a) : (b);
}
inline long long MIN3(long long a, long long b, long long c) {
  return (a) < (b) ? ((a) < (c) ? (a) : (c)) : ((b) < (c) ? (b) : (c));
}
using namespace std;
int main() {
  int i;
  long long n, m;
  long long p[200005];
  cin >> n >> m;
  map<long long, long long> left, right;
  map<long long, long long>::iterator it, it1;
  long long mpos;
  for (i = 0; i < n; i++) {
    cin >> p[i];
    if (p[i] == m) mpos = i;
  }
  long long less = 0, more = 0;
  for (i = mpos - 1; i >= 0; i--) {
    if (p[i] < m)
      less++;
    else
      more++;
    left[less - more]++;
  }
  less = 0, more = 0;
  for (i = mpos + 1; i < n; i++) {
    if (p[i] < m)
      less++;
    else
      more++;
    right[less - more]++;
  }
  long long ans = 1;
  for (i = 0; i >= -1; i--) {
    if (left.find(i) != left.end()) ans += left[i];
    if (right.find(i) != right.end()) ans += right[i];
  }
  for (it = right.begin(); it != right.end(); it++) {
    for (i = 0; i >= -1; i--) {
      it1 = left.find(-it->first + i);
      if (it1 != left.end()) ans += it->second * it1->second;
    }
  }
  cout << ans;
}

```