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>
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse4")
using namespace std;
inline long long in() {
  long long x;
  scanf("%lld", &x);
  return x;
}
const long long N = 1e6 + 100;
long long offset = 2e5 + 2;
long long tree[N], a[N], b[N], n;
void upd(long long id, long long x) {
  while (id < N) {
    tree[id] += x;
    id += (id & -id);
  }
}
long long get(long long id) {
  long long ans = 0;
  while (id >= 1) {
    ans += tree[id];
    id -= (id & -id);
  }
  return ans;
}
long long g(long long m) {
  for (long long i = 1; i < N; i++) tree[i] = 0;
  for (long long i = 1; i <= n; i++)
    if (a[i] < m)
      b[i] = -1;
    else
      b[i] = 1;
  for (long long i = 1; i <= n; i++) b[i] += b[i - 1];
  for (long long i = 0; i <= n; i++) b[i] += offset;
  long long ct = 0;
  for (long long i = 0; i <= n; i++) {
    ct += get(b[i]);
    upd(b[i] + 1, 1);
  }
  return ct;
}
int32_t main() {
  long long m;
  cin >> n >> m;
  for (long long i = 1; i <= n; i++) a[i] = in();
  cout << g(m) - g(m + 1) << "\n";
}
