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 max_n = 400222, inf = 1000111222;
int n, m, a[max_n], t[max_n];
void inc(int i) {
  i += max_n / 2;
  for (; i < max_n; i = (i | (i + 1))) {
    ++t[i];
  }
}
int get(int r) {
  r += max_n / 2;
  int res = 0;
  for (; r >= 0; r = (r & (r + 1)) - 1) {
    res += t[r];
  }
  return res;
}
long long solve(int r) {
  memset(t, 0, sizeof(t));
  inc(0);
  int sum = 0;
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    if (a[i] <= r) {
      --sum;
    } else {
      ++sum;
    }
    ans += get(-sum);
    inc(-sum);
  }
  return ans;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
  }
  cout << solve(m) - solve(m - 1) << "\n";
  return 0;
}
