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;
long long n, i, j, k, l, sum = 0, flag = 0, t, a[200008], b[200007], ans = 0;
string s;
int BIT[2][400050];
void update(int id, long long first, long long val) {
  first += 200005;
  for (; first <= 400050; first += (first & -first)) BIT[id][first] += val;
}
long long query(int id, long long first) {
  first += 200005;
  long long res = 0;
  for (; first > 0; first -= (first & -first)) res += BIT[id][first];
  return res;
}
long long solve(int k) {
  memset(BIT, 0, sizeof BIT);
  b[0] = 0;
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    b[i] = (a[i] >= k ? 1 : -1);
    b[i] += b[i - 1];
  }
  update(0, 0, 1);
  for (int i = 1; i <= n; i++) {
    ans += query(!(i & 1), b[i]) + query((i & 1), b[i] - 1);
    update(i & 1, b[i], 1);
  }
  return ans;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  cout << solve(k) - solve(k + 1) << '\n';
  return 0;
}
