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;
class SegmentTree {
 public:
  SegmentTree(int count) {
    n = count;
    data.assign(2 * n, 0);
  }
  SegmentTree(std::vector<int> const &values) {
    n = values.size();
    data.resize(2 * n);
    std::copy(values.begin(), values.end(), &data[0] + n);
    for (int idx = n - 1; idx > 0; idx--)
      data[idx] = data[2 * idx] + data[2 * idx + 1];
  }
  void update(int idx, int add) {
    idx += n;
    data[idx] += add;
    while (idx > 1) {
      idx /= 2;
      data[idx] = data[2 * idx] + data[2 * idx + 1];
    }
  }
  int sum(int left, int right = -1) {
    if (right == -1) right = n;
    int ret = 0;
    left += n;
    right += n;
    while (left < right) {
      if (left & 1) ret += data[left++];
      if (right & 1) ret += data[--right];
      left >>= 1;
      right >>= 1;
    }
    return ret;
  }

 private:
  int n;
  std::vector<int> data;
};
long long f(vector<int> v, int m) {
  long long res = 0;
  int n = v.size();
  SegmentTree st(n * 2 + 1);
  int d = 0;
  for (int i = 0; i < n; i++) {
    if (v[i] >= m)
      d++;
    else
      d--;
    st.update(d + n, 1);
  }
  d = 0;
  for (int i = 0; i < n; i++) {
    res += st.sum(1 + d + n);
    if (v[i] >= m)
      d++;
    else
      d--;
    st.update(d + n, -1);
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  int m;
  cin >> m;
  vector<int> v(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i];
  }
  cout << f(v, m) - f(v, m + 1) << '\n';
}
