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 = 2e5 + 500;
const int MOD = 1e9 + 7;
long long a[N], sum[N], tree[N * 2 + 2];
long long n, m;
long long query(long long pos) {
  long long res = 0;
  for (long long i = pos; i > 0; i -= i & (-i)) res += tree[i];
  return res;
}
void update(long long pos) {
  for (long long i = pos; i <= 2 * N; i += i & (-i)) tree[i] += 1;
}
long long x[N], y[N];
long long run(long long m) {
  memset(tree, 0, sizeof tree);
  memset(x, 0, sizeof x);
  memset(y, 0, sizeof y);
  memset(sum, 0, sizeof sum);
  long long res = 0;
  long long small = 0, big = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] > m)
      big++, x[i] = big, y[i] = y[i - 1];
    else
      small++, x[i] = x[i - 1], y[i] = small;
  }
  for (int i = 1; i <= n; i++) sum[i] = y[i] - x[i];
  update(n + 50);
  for (int r = 1; r <= n; r++) {
    res += query(sum[r] + n + 50);
    update(sum[r] + n + 50);
  }
  return res;
}
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i];
  cout << run(m) - run(m - 1) << endl;
  return 0;
}
