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 read() {
  char c;
  int num, f = 1;
  while (c = (char)getchar(), !isdigit(c))
    if (c == '-') f = -1;
  num = (int)(c - '0');
  while (c = (char)getchar(), isdigit(c)) num = num * 10 + (int)(c - '0');
  return num * f;
}
int n, m;
vector<int> a;
long long greaterCount(int m) {
  vector<int> s(2 * n + 1);
  int sum = n;
  long long result = 0;
  s[sum] = 1;
  long long add = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] < m)
      sum--, add -= s[sum];
    else
      add += s[sum], sum++;
    result += add;
    s[sum]++;
  }
  return result;
}
void work() {
  cin >> n >> m;
  a = vector<int>(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  cout << greaterCount(m) - greaterCount(m + 1) << endl;
  return;
}
signed main() {
  int T = 1;
  for (int Case = 1; Case <= T; Case++) {
    work();
  }
  return 0;
}
