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;
template <class T>
struct FenwickTree {
  std::vector<T> in;
  int N;
  FenwickTree(int N) : N(N), in(N) {}
  void add(int at, T by) {
    for (int i = at; i < N; i += (i & -i)) {
      in[i] += by;
    }
  }
  T query(int at) {
    T sum = 0;
    for (int i = at; i; i -= (i & -i)) {
      sum += in[i];
    }
    return sum;
  }
};
FenwickTree<int> ft(400005);
struct Query {
  int x, y;
  int id;
};
int ans[200005];
vector<Query> cdq(vector<Query> vec) {
  if (vec.size() <= 1) return vec;
  int m = vec.size() / 2;
  auto left = cdq({vec.begin(), vec.begin() + m});
  auto right = cdq({vec.begin() + m, vec.end()});
  vector<Query> ret;
  int h1 = 0, h2 = 0;
  while (h1 < left.size() || h2 < right.size()) {
    if (h2 == right.size() || (h1 < left.size() && left[h1].x < right[h2].x)) {
      ft.add(left[h1].y, 1);
      ret.push_back(left[h1++]);
    } else {
      ans[right[h2].id] += ft.query(right[h2].y);
      ret.push_back(right[h2++]);
    }
  }
  for (auto q : left) {
    ft.add(q.y, -1);
  }
  return ret;
}
int main(int argc, char *argv[]) {
  std::ios_base::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  vector<int> a(n + 1), b(n + 1), c(n + 1);
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    b[i] = b[i - 1] + (a[i] <= m);
    c[i] = c[i - 1] + (a[i] < m);
  }
  vector<Query> vec;
  for (int i = 0; i <= n; i++) {
    vec.push_back({-(c[i] * 2 - i), b[i] * 2 - i + 200000, i});
  }
  cdq(vec);
  long long sum = 0;
  for (int i = 0; i <= n; i++) {
    sum += ans[i];
  }
  cout << sum << endl;
  return 0;
}
