You are given an integer sequence a_1, a_2, ..., a_n.

Find the number of pairs of indices (l, r) (1 ≤ l ≤ r ≤ n) such that the value of median of a_l, a_{l+1}, ..., a_r is exactly the given number m.

The median of a sequence is the value of an 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 median of a_l, a_{l+1}, ..., a_r is exactly the given number m.

Input

The first line contains integers n and m (1 ≤ n,m ≤ 2⋅10^5) — the length of the given sequence and the required value of the median.

The second line contains an integer sequence a_1, a_2, ..., a_n (1 ≤ a_i ≤ 2⋅10^5).

Output

Print the required number.

Examples

Input

5 4
1 4 5 60 4


Output

8


Input

3 1
1 1 1


Output

6


Input

15 2
1 2 3 1 2 3 1 2 3 1 2 3 1 2 3


Output

97

Note

In the first example, the suitable pairs of indices are: (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 5), (4, 5) and (5, 5).
#include <bits/stdc++.h>
using namespace std;
const int maxn = 5e5 + 100;
int arr[maxn];
int zlicz[2 * maxn];
void add(int it) {
  while (it < 2 * maxn) {
    zlicz[it]++;
    it += it & (-it);
  }
}
int query(int r) {
  int ret = 0;
  while (r > 0) {
    ret += zlicz[r];
    r -= r & (-r);
  }
  return ret;
}
long long cunt(int n, int m) {
  memset(zlicz, 0, sizeof(zlicz));
  int sum = maxn;
  int dod = 0;
  long long ret = 0;
  add(sum);
  for (int i = 1; i <= n; i++) {
    if (arr[i] >= m) {
      ret += query(sum);
      sum++;
    } else {
      sum--;
      ret += query(sum - 1);
    }
    add(sum);
  }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> arr[i];
  cout << cunt(n, m) - cunt(n, m + 1) << "\n";
}
