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 long long INF = 2e9;
const long long MOD = 1e9 + 7;
struct FenvikTree {
  vector<long long> t;
  FenvikTree(long long n) { t.resize(n + 1); }
  long long get(long long i) {
    i++;
    if (!i) return 0;
    long long res = 0;
    for (; i > 0; i -= (i & (-i))) {
      res += t[i];
    }
    return res;
  }
  void update(long long i) {
    i++;
    for (; i < t.size(); i += (i & (-i))) {
      t[i]++;
    }
  }
};
long long get(long long n, long long m, vector<long long> a) {
  vector<long long> pref(n + 1);
  for (long long i = 1; i <= n; i++) {
    pref[i] = pref[i - 1] + (a[i - 1] >= m ? 1 : -1);
  }
  FenvikTree tree(2 * n + 1);
  long long ans = 0;
  for (long long i = 0; i <= n; i++) {
    ans += tree.get(pref[i] + n - 1);
    tree.update(pref[i] + n);
  }
  return ans;
}
void solve() {
  long long n, m;
  cin >> n >> m;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  cout << get(n, m, a) - get(n, m + 1, a);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cout.precision(12);
  solve();
}
