#include <bits/stdc++.h>
using namespace std;
const long long maxn = 1e6 + 5;
const long long maxint = 1e9 + 7;
long long n, k, ans = 0, a[maxn], b[maxn], BIT[1000005];
vector<long long> zip;
inline void Update(long long index, long long val) {
  index++;
  while (index <= zip.size() + 1) {
    BIT[index] += val;
    index += index & -index;
  }
}
inline long long Query(long long index) {
  long long ret = 0;
  index++;
  while (index > 0) {
    ret += BIT[index];
    index -= index & -index;
  }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  scanf("%lld%lld", &n, &k);
  for (long long i = 1; i <= n; i++) scanf("%lld", a + i);
  b[0] = 0;
  for (long long i = 1; i <= n; i++) {
    b[i] = b[i - 1] + (a[i] >= k);
    zip.push_back((b[i] << 1) - i - 1);
    zip.push_back((b[i - 1] << 1) - i);
  }
  sort(zip.begin(), zip.end());
  zip.erase(unique(zip.begin(), zip.end()), zip.end());
  for (long long i = 1; i <= n; i++) {
    long long prePos =
        lower_bound(zip.begin(), zip.end(), (b[i - 1] << 1) - i) - zip.begin();
    Update(prePos, 1);
    long long pos =
        lower_bound(zip.begin(), zip.end(), (b[i] << 1) - i - 1) - zip.begin();
    ans += Query(pos - 1);
  }
  long long fi = ans;
  ans = 0;
  k++;
  b[0] = 0;
  zip.clear();
  memset(BIT, 0, sizeof BIT);
  for (long long i = 1; i <= n; i++) {
    b[i] = b[i - 1] + (a[i] >= k);
    zip.push_back((b[i] << 1) - i - 1);
    zip.push_back((b[i - 1] << 1) - i);
  }
  sort(zip.begin(), zip.end());
  zip.erase(unique(zip.begin(), zip.end()), zip.end());
  for (long long i = 1; i <= n; i++) {
    long long prePos =
        lower_bound(zip.begin(), zip.end(), (b[i - 1] << 1) - i) - zip.begin();
    Update(prePos, 1);
    long long pos =
        lower_bound(zip.begin(), zip.end(), (b[i] << 1) - i - 1) - zip.begin();
    ans += Query(pos - 1);
  }
  fi -= ans;
  cout << fi;
  return 0;
}

### Prompt

Create a solution in cpp for 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;
const long long maxn = 1e6 + 5;
const long long maxint = 1e9 + 7;
long long n, k, ans = 0, a[maxn], b[maxn], BIT[1000005];
vector<long long> zip;
inline void Update(long long index, long long val) {
  index++;
  while (index <= zip.size() + 1) {
    BIT[index] += val;
    index += index & -index;
  }
}
inline long long Query(long long index) {
  long long ret = 0;
  index++;
  while (index > 0) {
    ret += BIT[index];
    index -= index & -index;
  }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  scanf("%lld%lld", &n, &k);
  for (long long i = 1; i <= n; i++) scanf("%lld", a + i);
  b[0] = 0;
  for (long long i = 1; i <= n; i++) {
    b[i] = b[i - 1] + (a[i] >= k);
    zip.push_back((b[i] << 1) - i - 1);
    zip.push_back((b[i - 1] << 1) - i);
  }
  sort(zip.begin(), zip.end());
  zip.erase(unique(zip.begin(), zip.end()), zip.end());
  for (long long i = 1; i <= n; i++) {
    long long prePos =
        lower_bound(zip.begin(), zip.end(), (b[i - 1] << 1) - i) - zip.begin();
    Update(prePos, 1);
    long long pos =
        lower_bound(zip.begin(), zip.end(), (b[i] << 1) - i - 1) - zip.begin();
    ans += Query(pos - 1);
  }
  long long fi = ans;
  ans = 0;
  k++;
  b[0] = 0;
  zip.clear();
  memset(BIT, 0, sizeof BIT);
  for (long long i = 1; i <= n; i++) {
    b[i] = b[i - 1] + (a[i] >= k);
    zip.push_back((b[i] << 1) - i - 1);
    zip.push_back((b[i - 1] << 1) - i);
  }
  sort(zip.begin(), zip.end());
  zip.erase(unique(zip.begin(), zip.end()), zip.end());
  for (long long i = 1; i <= n; i++) {
    long long prePos =
        lower_bound(zip.begin(), zip.end(), (b[i - 1] << 1) - i) - zip.begin();
    Update(prePos, 1);
    long long pos =
        lower_bound(zip.begin(), zip.end(), (b[i] << 1) - i - 1) - zip.begin();
    ans += Query(pos - 1);
  }
  fi -= ans;
  cout << fi;
  return 0;
}

```