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>
const int MAXN = 2e5;
using namespace std;
inline int read() {
  int x = 0, w = 1;
  char c = ' ';
  while (c < '0' || c > '9') {
    c = getchar();
    if (c == '-') w = -1;
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  return x * w;
}
int n, m, a[MAXN + 5];
int cnt[MAXN * 2 + 5];
void init() {
  n = read();
  m = read();
  for (int i = 1; i <= n; ++i) a[i] = read();
}
inline long long calc(int x) {
  memset(cnt, 0, sizeof(cnt));
  long long ret = 0;
  int cur = 0, sum = 0;
  cnt[0 + MAXN] = 1;
  for (int i = 1; i <= n; ++i) {
    if (a[i] >= x) {
      sum += cnt[cur + MAXN];
      ++cur;
    } else {
      --cur;
      sum -= cnt[cur + MAXN];
    }
    ret += sum;
    ++cnt[cur + MAXN];
  }
  return ret;
}
int main() {
  init();
  printf("%I64d\n", calc(m) - calc(m + 1));
  return 0;
}
