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;
int n, m, a[200010];
struct fenwick {
  vector<int> F;
  int N;
  void init(int n) {
    N = n;
    F.clear();
    F.resize(N + 1);
  }
  void add(int k, int v) {
    if (k == 0) return;
    while (k <= N) {
      F[k] += v;
      k += k & -k;
    }
  }
  int sum(int k) {
    int s = 0;
    while (k >= 1) {
      s += F[k];
      k -= k & -k;
    }
    return s;
  }
} f;
long long calc(int k) {
  int h = 0, l = 0;
  f.init(200010 * 2);
  for (int i = 0; i < n; ++i) {
    l += a[i] < k;
    h += a[i] >= k;
    f.add(h - l + 200010, 1);
  }
  long long ret = 0;
  int cur = 0;
  h = l = 0;
  for (int i = 0; i < n; ++i) {
    ret += f.sum(2 * 200010) - f.sum(cur + 200010);
    l += a[i] < k;
    h += a[i] >= k;
    f.add(h - l + 200010, -1);
    cur += (a[i] >= k) - (a[i] < k);
  }
  assert(f.sum(2 * 200010) == 0);
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> m;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  cout << calc(m) - calc(m + 1) << endl;
  return 0;
}
