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;
void _out() { cerr << '\n'; }
template <typename H, typename... T>
void _out(H h, T... t) {
  cerr << ' ' << h;
  _out(t...);
}
const long long N = 1e6 + 100;
long long tr[N];
long long lowbit(long long x) { return x & -x; }
long long sum(long long x) {
  long long res = 0;
  for (long long i = x; i; i -= lowbit(i)) res += tr[i];
  return res;
}
void add(long long x, long long c) {
  for (long long i = x; i <= 1e6; i += lowbit(i)) {
    tr[i] += c;
  }
}
void solve() {
  long long n, m;
  cin >> n >> m;
  vector<long long> a(n + 1);
  for (long long i = 1; i <= n; i++) cin >> a[i];
  auto get = [&](long long x) {
    long long res = 0;
    memset(tr, 0, sizeof tr);
    vector<long long> b(n + 1);
    add(n + 1, 1);
    for (long long i = 1; i <= n; i++) {
      b[i] = a[i] >= x ? 1 : -1;
      b[i] += b[i - 1];
    }
    for (long long i = 1; i <= n; i++) {
      res += sum(b[i] + n);
      add(b[i] + n + 1, 1);
    }
    return res;
  };
  cout << get(m) - get(m + 1);
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  solve();
}
