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 + 5;
int a[N], n;
int tree[N << 3];
void updata(int root, int l, int r, int pos, int v) {
  if (l == r) {
    tree[root] += v;
    return;
  }
  int mid = (l + r) / 2;
  if (pos <= mid)
    updata(root << 1, l, mid, pos, v);
  else
    updata(root << 1 | 1, mid + 1, r, pos, v);
  tree[root] = tree[root << 1] + tree[root << 1 | 1];
}
int query(int root, int l, int r, int ul, int ur) {
  if (ul <= l && r <= ur) return tree[root];
  int mid = (l + r) / 2;
  int ret = 0;
  if (mid >= ul) ret += query(root << 1, l, mid, ul, ur);
  if (mid < ur) ret += query(root << 1 | 1, mid + 1, r, ul, ur);
  return ret;
}
long long solve(int m) {
  int up = 0, down = 0;
  memset(tree, 0, sizeof(tree));
  long long ans = 0;
  updata(1, 1, 2 * N, N, 1);
  for (int i = 1; i <= n; i++) {
    if (a[i] >= m)
      up++;
    else
      down++;
    ans += query(1, 1, 2 * N, 1, up - down + N - 1);
    updata(1, 1, 2 * N, up - down + N, 1);
  }
  return ans;
}
int main() {
  int m;
  while (~scanf("%d %d", &n, &m)) {
    for (int i = 1; i <= n; i++) scanf("%d", a + i);
    printf("%lld\n", solve(m) - solve(m + 1));
  }
  return 0;
}
