#include <bits/stdc++.h>
using namespace std;
void doRoutine() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
const long long MAXN = 200005;
long long a[MAXN];
long long n, m;
long long arr[MAXN];
long long solve(long long l, long long r) {
  if (l + 1 == r) {
    if (arr[l] == 1) {
      return 1;
    } else {
      return 0;
    }
  }
  long long m = (l + r) / 2;
  long long res = 0;
  res += solve(l, m);
  res += solve(m, r);
  vector<long long> cnt((r - l) * 2 + 5);
  long long sh = (r - l);
  long long cur = 0;
  for (long long i = m - 1; i >= l; --i) {
    cur += arr[i];
    ++cnt[cur + sh];
  }
  vector<long long> suf_cnt(cnt.size() + 1);
  for (long long i = 1; i < suf_cnt.size(); ++i) {
    suf_cnt[i] = suf_cnt[i - 1] + cnt[(long long)cnt.size() - i];
  }
  cur = 0;
  for (long long i = m; i < r; ++i) {
    cur += arr[i];
    res += suf_cnt[(long long)cnt.size() - (sh - cur) - 1];
  }
  return res;
}
long long f(long long k) {
  for (long long i = 0; i < n; ++i) {
    if (a[i] >= k) {
      arr[i] = 1;
    } else {
      arr[i] = -1;
    }
  }
  return solve(0, n);
}
signed main() {
  doRoutine();
  cin >> n >> m;
  for (long long i = 0; i < n; ++i) {
    cin >> a[i];
  }
  long long ans1 = f(m + 1), ans2 = f(m);
  cout << ans2 - ans1 << endl;
  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;
void doRoutine() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
const long long MAXN = 200005;
long long a[MAXN];
long long n, m;
long long arr[MAXN];
long long solve(long long l, long long r) {
  if (l + 1 == r) {
    if (arr[l] == 1) {
      return 1;
    } else {
      return 0;
    }
  }
  long long m = (l + r) / 2;
  long long res = 0;
  res += solve(l, m);
  res += solve(m, r);
  vector<long long> cnt((r - l) * 2 + 5);
  long long sh = (r - l);
  long long cur = 0;
  for (long long i = m - 1; i >= l; --i) {
    cur += arr[i];
    ++cnt[cur + sh];
  }
  vector<long long> suf_cnt(cnt.size() + 1);
  for (long long i = 1; i < suf_cnt.size(); ++i) {
    suf_cnt[i] = suf_cnt[i - 1] + cnt[(long long)cnt.size() - i];
  }
  cur = 0;
  for (long long i = m; i < r; ++i) {
    cur += arr[i];
    res += suf_cnt[(long long)cnt.size() - (sh - cur) - 1];
  }
  return res;
}
long long f(long long k) {
  for (long long i = 0; i < n; ++i) {
    if (a[i] >= k) {
      arr[i] = 1;
    } else {
      arr[i] = -1;
    }
  }
  return solve(0, n);
}
signed main() {
  doRoutine();
  cin >> n >> m;
  for (long long i = 0; i < n; ++i) {
    cin >> a[i];
  }
  long long ans1 = f(m + 1), ans2 = f(m);
  cout << ans2 - ans1 << endl;
  return 0;
}

```