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 <typename T>
inline T MIN(const T &a, const T &b) {
  return a < b ? a : b;
}
template <typename T>
inline T MAX(const T &a, const T &b) {
  return a > b ? a : b;
}
template <typename T, typename... Args>
inline T MIN(const T &a, const T &b, Args... args) {
  return MIN(MIN(a, b), args...);
}
template <typename T, typename... Args>
inline T MAX(const T &a, const T &b, Args... args) {
  return MAX(MAX(a, b), args...);
}
const double eps = 1e-8;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const int dir[4][2] = {-1, 0, 1, 0, 0, -1, 0, 1};
long long bit[400010], a[200010];
void edit(long long pos) {
  for (; pos < 400010; pos += pos & (-pos)) bit[pos]++;
}
long long query(long long pos) {
  long long res = 0;
  for (; pos; pos -= pos & (-pos)) res += bit[pos];
  return res;
}
long long solve(int n, long long m) {
  memset(bit, 0, sizeof bit);
  int sum = n + 1;
  long long res = 0;
  edit(sum);
  for (int i = 0; i < n; i++) {
    sum += (a[i] >= m ? 1 : -1);
    res += query(sum - 1);
    edit(sum);
  }
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  long long m;
  cin >> n >> m;
  for (int i = 0; i < n; i++) cin >> a[i];
  cout << solve(n, m) - solve(n, m + 1) << '\n';
  return 0;
}
