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 long long mod = 998244353;
inline long long add(long long v1, long long v2) {
  v1 += v2;
  if (v1 >= mod) v1 -= mod;
  if (v1 < 0) v1 += mod;
  return v1;
}
inline long long mul(long long v1, long long v2) { return v1 * v2 % mod; }
inline long long mypow(long long v, long long t) {
  long long res = 1;
  while (t) {
    if (t & 1) res = res * v % mod;
    v = v * v % mod;
    t >>= 1;
  }
  return res;
}
inline void bye() {
  puts("");
  exit(0);
}
const int N = 404040;
int sum[N];
int a[N];
long long sol(int n, int m) {
  for (int i = 0; i < N; i++) sum[i] = 0;
  long long res = 0;
  long long pos = n + 3;
  long long cur = 0;
  sum[pos] = 1;
  for (int i = 1; i <= n; i++) {
    if (a[i] < m) {
      pos--;
      cur -= sum[pos];
    } else {
      cur += sum[pos];
      pos++;
    }
    sum[pos]++;
    res += cur;
  }
  return res;
}
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  long long ans = sol(n, m) - sol(n, m + 1);
  printf("%lld\n", ans);
  return 0;
}
