#include <bits/stdc++.h>
using namespace std;
long long choose(long long n, long long k);
bool isprime(long long n);
long long power(long long x, long long y, long long p);
int maxi(int Count, ...);
int mini(int Count, ...);
map<long long, long long> mpp;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, m;
  cin >> n >> m;
  long long a[n], id;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] == m) id = i;
  }
  long long j = id - 1, cnt1 = 0, cnt2 = 0, ans = 1;
  while (j >= 0) {
    if (a[j] > m)
      cnt1++;
    else if (a[j] < m)
      cnt2++;
    if (cnt1 == cnt2 || cnt1 == cnt2 + 1) ans++;
    j--;
  }
  j = id + 1, cnt1 = 0, cnt2 = 0;
  while (j < n) {
    if (a[j] > m)
      cnt1++;
    else if (a[j] < m)
      cnt2++;
    if (cnt1 == cnt2 || cnt1 == cnt2 + 1) ans++;
    j++;
  }
  long long temp1 = 0, temp2 = 0;
  cnt1 = 0, cnt2 = 0;
  for (int k = id + 1; k < n; k++) {
    if (a[k] > m)
      cnt1++;
    else
      cnt1--;
    mpp[cnt1]++;
  }
  cnt1 = 0, cnt2 = 0;
  for (int i = id - 1; i > -1; i--) {
    if (a[i] > m)
      temp1++;
    else
      temp1--;
    if (mpp[-1 * temp1] > 0) ans += mpp[-1 * temp1];
    if (mpp[-1 * temp1 + 1] > 0) ans += mpp[-1 * temp1 + 1];
  }
  cout << ans << endl;
  return 0;
}
long long choose(long long n, long long k) {
  if (k == 0) return 1;
  return (n * choose(n - 1, k - 1)) / k;
}
bool isprime(long long n) {
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
int maxi(int Count, ...) {
  va_list Numbers;
  va_start(Numbers, Count);
  int ma = INT_MIN;
  for (int i = 0; i < Count; i++) ma = max(ma, va_arg(Numbers, int));
  va_end(Numbers);
  return ma;
}
int mini(int Count, ...) {
  va_list Numbers;
  va_start(Numbers, Count);
  int mi = INT_MAX;
  for (int i = 0; i < Count; i++) mi = min(mi, va_arg(Numbers, int));
  va_end(Numbers);
  return mi;
}

### 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 choose(long long n, long long k);
bool isprime(long long n);
long long power(long long x, long long y, long long p);
int maxi(int Count, ...);
int mini(int Count, ...);
map<long long, long long> mpp;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, m;
  cin >> n >> m;
  long long a[n], id;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] == m) id = i;
  }
  long long j = id - 1, cnt1 = 0, cnt2 = 0, ans = 1;
  while (j >= 0) {
    if (a[j] > m)
      cnt1++;
    else if (a[j] < m)
      cnt2++;
    if (cnt1 == cnt2 || cnt1 == cnt2 + 1) ans++;
    j--;
  }
  j = id + 1, cnt1 = 0, cnt2 = 0;
  while (j < n) {
    if (a[j] > m)
      cnt1++;
    else if (a[j] < m)
      cnt2++;
    if (cnt1 == cnt2 || cnt1 == cnt2 + 1) ans++;
    j++;
  }
  long long temp1 = 0, temp2 = 0;
  cnt1 = 0, cnt2 = 0;
  for (int k = id + 1; k < n; k++) {
    if (a[k] > m)
      cnt1++;
    else
      cnt1--;
    mpp[cnt1]++;
  }
  cnt1 = 0, cnt2 = 0;
  for (int i = id - 1; i > -1; i--) {
    if (a[i] > m)
      temp1++;
    else
      temp1--;
    if (mpp[-1 * temp1] > 0) ans += mpp[-1 * temp1];
    if (mpp[-1 * temp1 + 1] > 0) ans += mpp[-1 * temp1 + 1];
  }
  cout << ans << endl;
  return 0;
}
long long choose(long long n, long long k) {
  if (k == 0) return 1;
  return (n * choose(n - 1, k - 1)) / k;
}
bool isprime(long long n) {
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
int maxi(int Count, ...) {
  va_list Numbers;
  va_start(Numbers, Count);
  int ma = INT_MIN;
  for (int i = 0; i < Count; i++) ma = max(ma, va_arg(Numbers, int));
  va_end(Numbers);
  return ma;
}
int mini(int Count, ...) {
  va_list Numbers;
  va_start(Numbers, Count);
  int mi = INT_MAX;
  for (int i = 0; i < Count; i++) mi = min(mi, va_arg(Numbers, int));
  va_end(Numbers);
  return mi;
}

```