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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long tre[3 * 400005];
void init(long long n, long long l, long long r) {
  if (l == r) {
    tre[n] = 0;
    return;
  }
  long long mid = (l + r) / 2;
  init(2 * n, l, mid);
  init(2 * n + 1, mid + 1, r);
  tre[n] = tre[2 * n] + tre[2 * n + 1];
}
void update(long long n, long long l, long long r, long long i) {
  if (i < l || i > r) return;
  if (i >= r && i <= l) {
    tre[n]++;
    return;
  }
  long long mid = (l + r) / 2;
  update(2 * n, l, mid, i);
  update(2 * n + 1, mid + 1, r, i);
  tre[n] = tre[2 * n] + tre[2 * n + 1];
};
long long query(long long n, long long l, long long r, long long i,
                long long j) {
  if (i > r || j < l) return 0;
  if (i <= l && j >= r) {
    return tre[n];
  }
  long long mid = (l + r) / 2;
  return query(2 * n, l, mid, i, j) + query(2 * n + 1, mid + 1, r, i, j);
}
long long arr[400005], n;
long long f(long long x) {
  long long md = 200000;
  init(1, 0, 2 * md);
  update(1, 0, 2 * md, md);
  long long cur = 0;
  long long ans = 0;
  for (long long i = 0; i < n; i++) {
    if (arr[i] < x)
      cur--;
    else
      cur++;
    ans += query(1, 0, 2 * md, 0, md + cur - 1);
    update(1, 0, 2 * md, md + cur);
  }
  return ans;
}
void solve() {
  long long i, j, k, l, r, m, a, b, c, d, x, y, z, t, p;
  cin >> n >> x;
  for (i = 0; i < n; i++) {
    cin >> arr[i];
  }
  cout << f(x) - f(x + 1) << endl;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long t = 1;
  while (t--) solve();
  return 0;
}
