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 maxn = 4e5 + 10;
int n, m, a[maxn], tree[maxn * 4], b[maxn];
void update(int p, int l, int r, int x) {
  if (l == r) {
    tree[p]++;
    return;
  }
  int m = (l + r) / 2;
  if (x <= m)
    update(p * 2, l, m, x);
  else
    update(p * 2 + 1, m + 1, r, x);
  tree[p] = tree[p * 2] + tree[p * 2 + 1];
}
long long query(int p, int l, int r, int x, int y) {
  if (l >= x && r <= y) {
    return tree[p];
  }
  long long ans = 0, m = (l + r) / 2;
  if (x <= m) ans += query(p * 2, l, m, x, y);
  if (y > m) ans += query(p * 2 + 1, m + 1, r, x, y);
  return ans;
}
long long solve(int m) {
  long long ans = 0, sum = n + 2;
  memset(tree, 0, sizeof(tree));
  for (int i = 1; i <= n; i++) {
    if (a[i] >= m)
      b[i] = 1;
    else
      b[i] = -1;
  }
  update(1, 1, maxn, sum);
  for (int i = 1; i <= n; i++) {
    sum += b[i];
    ans += query(1, 1, maxn, 1, sum - 1);
    update(1, 1, maxn, sum);
  }
  return ans;
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  cout << solve(m) - solve(m + 1);
  return 0;
}
