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;
mt19937 rng(time(0));
uniform_int_distribution<int> uid(-1e9, 1e9);
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n, m;
  cin >> n >> m;
  vector<int> v(n);
  for (auto& i : v) cin >> i;
  map<int, int> mp, mp1;
  long long res = 0, s = 0, x = 0, s1 = 0, x1 = 0;
  mp[0] = 1;
  mp1[0] = 1;
  for (auto& i : v) {
    if (i < m)
      x -= mp[--s]++;
    else
      x += mp[s++], ++mp[s];
    if (i <= m)
      x1 -= mp1[--s1]++;
    else
      x1 += mp1[s1++], ++mp1[s1];
    res += x - x1;
  }
  cout << res;
}
