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 = (int)2e5 + 5;
const int MOD = (int)1e9 + 7;
int a[MAXN];
pair<int, int> p[MAXN];
int sum[MAXN], n, m;
void add(int x, int y = 1) {
  for (int i = x; i <= n + 1; i += i & (-i)) sum[i] += y;
}
int query(int x) {
  int re = 0;
  x = min(x, n + 1);
  for (int i = x; i; i -= i & (-i)) re += sum[i];
  return re;
}
long long solve1() {
  vector<int> ve;
  for (int i = 0; i <= n; i++) {
    ve.push_back(p[i].second);
  }
  sort(ve.begin(), ve.end());
  ve.erase(unique(ve.begin(), ve.end()), ve.end());
  sort(p, p + 1 + n);
  int stp = 0;
  long long re = 0;
  for (int i = 0; i <= n; i++) {
    int idx = upper_bound(ve.begin(), ve.end(), p[i].second - 1) - ve.begin();
    re += query(idx);
    p[i].second =
        lower_bound(ve.begin(), ve.end(), p[i].second) - ve.begin() + 1;
    add(p[i].second);
  }
  return re;
}
map<int, int> dp;
long long solve2() {
  int lst = 0;
  long long re = 0;
  for (int i = 1; i <= n + 1; i++) {
    if (i == n + 1 || a[i] == m) {
      dp.clear();
      int stp = 0;
      for (int j = lst + 1; j < i; j++) {
        dp[stp]++;
        if (a[j] > m)
          stp++;
        else
          stp--;
        re += dp[stp] + dp[stp - 1];
      }
      lst = i;
    }
  }
  return re;
}
int main() {
  scanf("%d%d", &n, &m);
  int x, y;
  x = y = 0;
  p[0] = make_pair(0, 0);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == m)
      y++;
    else if (a[i] > m)
      x++;
    else
      x--;
    p[i] = make_pair(y - x, x + y);
  }
  long long ans = solve1();
  printf("%lld\n", ans);
  return 0;
}
