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;
int cnt[200005], b[200005], n;
int a[400015];
int lowbit(int x) { return x & (-x); }
void modify(int x, int add) {
  while (x <= 400005) {
    a[x] += add;
    x += lowbit(x);
  }
}
int get_sum(int x) {
  int ret = 0;
  while (x != 0) {
    ret += a[x];
    x -= lowbit(x);
  }
  return ret;
}
long long slo(int x) {
  long long ans = 0;
  cnt[0] = 0;
  for (int i = 1; i <= n; ++i) cnt[i] = cnt[i - 1] + (b[i] >= x ? 1 : -1);
  for (int i = 0; i <= (n << 1); ++i) a[i] = 0;
  modify(0 + n + 1, 1);
  for (int i = 1; i <= n; ++i) {
    ans += get_sum(cnt[i] + n + 1 - 1);
    modify(cnt[i] + n + 1, 1);
  }
  return ans;
}
int main() {
  int m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d", b + i);
  printf("%I64d\n", slo(m) - slo(m + 1));
  return 0;
}
