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 = 4e5 + 200;
int a[maxn], c[maxn], b[maxn];
int n;
int sum(int x) {
  int ans = 0;
  while (x > 0) {
    ans += c[x];
    x -= (x & (-x));
  }
  return ans;
}
void add(int x) {
  while (x < maxn) {
    c[x] += 1;
    x += x & (-x);
  }
}
long long solve(int m) {
  memset(c, 0, sizeof(c));
  for (int i = 1; i <= n; i++) {
    b[i] = a[i] <= m ? 1 : -1;
    b[i] += b[i - 1];
  }
  add(n + 1);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    ans += sum(b[i] + n + 1);
    add(b[i] + n + 1);
  }
  return ans;
}
int main() {
  int m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  printf("%lld\n", solve(m) - solve(m - 1));
  return 0;
}
