#include <bits/stdc++.h>
using namespace std;

int main() {

  ios::sync_with_stdio(false);
  cout.tie(nullptr);
  cin.tie(nullptr);

  int n, k;
  cin >> n >> k;

  map<int, int> m;
  vector<int> a(n + 5), p(n + 5, 0);
  long long ans = 0, sum = 0;
  m[0] = 1;

  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    a[i] = (a[i] - 1 + k) % k;

    sum = (sum + a[i]) % k;
    if (i >= k) m[p[i-k]]--;
    // if (i >= k && a[i] == 0)  ans++;
    ans += m[sum];
    m[sum]++;
    p[i] = sum;
  }

  cout << ans << '\n';
}
### Prompt

Please create a solution in cpp to the following problem:

Given are a sequence of N positive integers A_1, A_2, \ldots, A_N, and a positive integer K.

Find the number of non-empty contiguous subsequences in A such that the remainder when dividing the sum of its elements by K is equal to the number of its elements. We consider two subsequences different if they are taken from different positions, even if they are equal sequences.

Constraints

* All values in input are integers.
* 1 \leq N \leq 2\times 10^5
* 1 \leq K \leq 10^9
* 1 \leq A_i \leq 10^9

Input

Input is given from Standard Input in the following format:


N K
A_1 A_2 \cdots A_N


Output

Print the number of subsequences that satisfy the condition.

Examples

Input

5 4
1 4 2 3 5


Output

4


Input

8 4
4 2 4 2 4 2 4 2


Output

7


Input

10 7
14 15 92 65 35 89 79 32 38 46


Output

8

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;

int main() {

  ios::sync_with_stdio(false);
  cout.tie(nullptr);
  cin.tie(nullptr);

  int n, k;
  cin >> n >> k;

  map<int, int> m;
  vector<int> a(n + 5), p(n + 5, 0);
  long long ans = 0, sum = 0;
  m[0] = 1;

  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    a[i] = (a[i] - 1 + k) % k;

    sum = (sum + a[i]) % k;
    if (i >= k) m[p[i-k]]--;
    // if (i >= k && a[i] == 0)  ans++;
    ans += m[sum];
    m[sum]++;
    p[i] = sum;
  }

  cout << ans << '\n';
}
```