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;
const int N = 4e5 + 5;
int id(int x) { return (x < 0) ? (200000 - x) : x; }
int ori(int x) { return (x > 200000) ? 200000 - x : x; }
vector<long long int> cnt(N, 0);
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<int> a(n, 0), x(n, 0);
  for (int j = 0; j < n; j++) {
    cin >> a[j];
    if (a[j] >= m)
      x[j] = 1;
    else
      x[j] = -1;
  }
  cnt[0] = 1;
  long long int pre = 0, ans = 0, cur_cnt = 1;
  for (int j = 0; j < n; j++) {
    pre += x[j];
    if (x[j] == 1)
      ;
    else
      cur_cnt -= (cnt[id(pre + 1)] + cnt[id(pre)]);
    cnt[id(pre)]++;
    ans += cur_cnt;
    cur_cnt += cnt[id(pre)];
  }
  cnt.clear();
  cnt.resize(N, 0);
  m++;
  for (int j = 0; j < n; j++)
    if (a[j] >= m)
      x[j] = 1;
    else
      x[j] = -1;
  cnt[0] = 1;
  pre = 0;
  cur_cnt = 1;
  for (int j = 0; j < n; j++) {
    pre += x[j];
    if (x[j] == 1)
      ;
    else
      cur_cnt -= (cnt[id(pre + 1)] + cnt[id(pre)]);
    cnt[id(pre)]++;
    ans -= cur_cnt;
    cur_cnt += cnt[id(pre)];
  }
  cout << ans << endl;
  return 0;
}
