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 INF = 0x3f3f3f3f;
const int MAXN = 400005;
const int MOD = 1e9 + 9;
int n;
int a[MAXN];
int pref[MAXN], sum[MAXN], s[MAXN];
long long grThan(int m) {
  long long ret = 0;
  for (int i = 0; i < 2 * n + 1; ++i) sum[i] = s[i] = 0;
  pref[0] = (a[0] < m ? n - 1 : n + 1);
  sum[n]++;
  s[n + 1]++;
  ret += s[pref[0]];
  sum[pref[0]]++;
  s[pref[0] + 1] = s[pref[0]] + sum[pref[0]];
  for (int i = 1; i < n; ++i) {
    pref[i] = pref[i - 1] + (a[i] < m ? -1 : 1);
    sum[pref[i]]++;
    ret += s[pref[i]];
    s[pref[i] + 1] = s[pref[i]] + sum[pref[i]];
  }
  return ret;
}
int main() {
  int m;
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
  long long ans = grThan(m) - grThan(m + 1);
  printf("%lld\n", ans);
  return 0;
}
