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

int main()
{
  cin.tie(0);
  cin.sync_with_stdio(0);

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

  vector < int >a(n);
  for (auto & x:a)
    cin >> x;
  a.insert(a.begin(), 0);

  ll ans = 0;
  map < int, int >cnt;
  for (int i = 1; i <= n; ++i)
    a[i] = (a[i] + a[i - 1] - 1) % k;

  for (int i = 0; i <= n; ++i) {
    if (i >= k)
      --cnt[a[i - k]];
    ans += cnt[a[i]]++;
  }

  cout << ans;
}

### Prompt

Create a solution in CPP for 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;
using ll = long long;

int main()
{
  cin.tie(0);
  cin.sync_with_stdio(0);

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

  vector < int >a(n);
  for (auto & x:a)
    cin >> x;
  a.insert(a.begin(), 0);

  ll ans = 0;
  map < int, int >cnt;
  for (int i = 1; i <= n; ++i)
    a[i] = (a[i] + a[i - 1] - 1) % k;

  for (int i = 0; i <= n; ++i) {
    if (i >= k)
      --cnt[a[i - k]];
    ans += cnt[a[i]]++;
  }

  cout << ans;
}

```