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 int maxn = (int)4e5 + 10;
const int mod = (int)998244353;
const int inf = (1 << 30) - 1;
int n;
int m;
int a[maxn];
int sumb[maxn];
int sumc[maxn];
int b[maxn];
int c[maxn];
map<int, int> t[maxn];
void add(int x, int y) {
  for (int i = x; i <= 2 * n; i = (i | (i + 1))) {
    for (int j = y; j <= 2 * n; j = (j | (j + 1))) {
      t[i][j]++;
    }
  }
}
int get(int x, int y) {
  int res = 0;
  for (int i = x; i >= 0; i = (i & (i + 1)) - 1) {
    for (int j = y; j >= 0; j = (j & (j + 1)) - 1) {
      if (t[i].count(j)) res += t[i][j];
    }
  }
  return res;
}
void solve() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] < m) {
      b[i] = -1;
      c[i] = -1;
    } else if (a[i] == m) {
      b[i] = -1;
      c[i] = 1;
    } else {
      b[i] = 1;
      c[i] = 1;
    }
    sumb[i] = sumb[i - 1] + b[i];
    sumc[i] = sumc[i - 1] + c[i];
  }
  int ind = 0;
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] == m) {
      while (ind < i) {
        add(sumb[ind] + n, sumc[ind] + n);
        ind++;
      }
    }
    ans += get(2 * n, sumc[i] + n - 1) - get(sumb[i] + n - 1, sumc[i] + n - 1);
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}
