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 P = (1000 * 1000 * 1000) + 7;
const int arrn = 2e5 + 3;
int n, m;
vector<int> in(arrn, 0);
long long comp(int m) {
  vector<int> temp;
  int count[2 * arrn] = {};
  temp.push_back(0);
  for (int i = 0; i < n; i++) {
    if (in[i] < m)
      temp.push_back(1);
    else
      temp.push_back(-1);
  }
  for (int i = 0; i < n; i++) temp[i + 1] += temp[i];
  count[0]++;
  int prev = 1;
  long long out = 0;
  for (int i = 0; i < n; i++) {
    int tempcount;
    if (temp[i + 1] < temp[i]) {
      if (temp[i] < 0)
        tempcount = count[arrn + temp[i]];
      else
        tempcount = count[temp[i]];
      prev = prev - tempcount;
      prev++;
    } else {
      if (temp[i + 1] < 0)
        tempcount = count[arrn + temp[i + 1]];
      else
        tempcount = count[temp[i + 1]];
      prev = prev + tempcount;
      prev++;
    }
    if (temp[i + 1] < 0)
      count[arrn + temp[i + 1]]++;
    else
      count[temp[i + 1]]++;
    out += (prev - 1);
  }
  return out;
}
void solve() {
  cin >> n >> m;
  for (int i = 0; i < n; i++) cin >> in[i];
  cout << comp(m + 1) - comp(m);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int T;
  T = 1;
  while (T--) {
    solve();
  }
  return 0;
}
