#include <bits/stdc++.h>
using namespace std;
const long long N = 2e5 + 5;
const long long offset = N + 1;
const long long MAX = N + offset + 1;
long long bit[MAX];
long long read(long long i) {
  long long ans = 0;
  i += offset;
  while (i > 0) {
    ans += bit[i];
    i -= (i & (-i));
  }
  return ans;
}
void update(long long i) {
  i += offset;
  while (i < MAX) {
    bit[i] += 1;
    i += (i & (-i));
  }
  return;
}
long long n, arr[N];
long long pref[N];
long long brute(long long m) {
  vector<pair<long long, long long> > subs;
  long long cnt = 0;
  for (long long i = 1; i <= n; ++i) {
    vector<long long> temp;
    for (long long j = i; j <= n; ++j) {
      temp.push_back(arr[j]);
      sort(temp.begin(), temp.end());
      long long idx = (temp.size() - 1) / 2;
      if (temp[idx] >= m) {
        ++cnt;
        subs.push_back({i, j});
      }
    }
  }
  return cnt;
}
long long calc(long long m) {
  for (long long i = 1; i < n + 1; ++i) {
    long long temp = (arr[i] >= m) ? 1 : -1;
    pref[i] = pref[i - 1] + temp;
  }
  memset(bit, 0, sizeof(bit));
  update(0);
  long long ans = 0;
  for (long long i = 1; i <= n; ++i) {
    long long got = read(pref[i] - 1);
    ans = ans + got;
    update(pref[i]);
  }
  return ans;
}
int32_t main() {
  long long m;
  cin >> n >> m;
  for (long long i = 1; i <= n; ++i) cin >> arr[i];
  long long ans = calc(m);
  ans -= calc(m + 1);
  cout << ans << endl;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
const long long N = 2e5 + 5;
const long long offset = N + 1;
const long long MAX = N + offset + 1;
long long bit[MAX];
long long read(long long i) {
  long long ans = 0;
  i += offset;
  while (i > 0) {
    ans += bit[i];
    i -= (i & (-i));
  }
  return ans;
}
void update(long long i) {
  i += offset;
  while (i < MAX) {
    bit[i] += 1;
    i += (i & (-i));
  }
  return;
}
long long n, arr[N];
long long pref[N];
long long brute(long long m) {
  vector<pair<long long, long long> > subs;
  long long cnt = 0;
  for (long long i = 1; i <= n; ++i) {
    vector<long long> temp;
    for (long long j = i; j <= n; ++j) {
      temp.push_back(arr[j]);
      sort(temp.begin(), temp.end());
      long long idx = (temp.size() - 1) / 2;
      if (temp[idx] >= m) {
        ++cnt;
        subs.push_back({i, j});
      }
    }
  }
  return cnt;
}
long long calc(long long m) {
  for (long long i = 1; i < n + 1; ++i) {
    long long temp = (arr[i] >= m) ? 1 : -1;
    pref[i] = pref[i - 1] + temp;
  }
  memset(bit, 0, sizeof(bit));
  update(0);
  long long ans = 0;
  for (long long i = 1; i <= n; ++i) {
    long long got = read(pref[i] - 1);
    ans = ans + got;
    update(pref[i]);
  }
  return ans;
}
int32_t main() {
  long long m;
  cin >> n >> m;
  for (long long i = 1; i <= n; ++i) cin >> arr[i];
  long long ans = calc(m);
  ans -= calc(m + 1);
  cout << ans << endl;
}

```