#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (n); i++)
using namespace std;
typedef long long ll;


int main()
{
  int n, k;
  cin >> n >> k;
  vector<ll> a(n + 1);
  rep(i, n)
  {
    int A;
    cin >> A;
    a[i+1] = a[i] + A - 1;
  }
  map<ll, int> mp;
  ll ans = 0;
  rep(i, n + 1)
  {
    if (i >= k)
      mp[a[i-k]%k]--;
    ans += mp[a[i]%k];
    mp[a[i]%k]++;
  }
  cout << ans << endl;
  return 0;
}
### Prompt

Please formulate a CPP solution 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>
#define rep(i,n) for (int i = 0; i < (n); i++)
using namespace std;
typedef long long ll;


int main()
{
  int n, k;
  cin >> n >> k;
  vector<ll> a(n + 1);
  rep(i, n)
  {
    int A;
    cin >> A;
    a[i+1] = a[i] + A - 1;
  }
  map<ll, int> mp;
  ll ans = 0;
  rep(i, n + 1)
  {
    if (i >= k)
      mp[a[i-k]%k]--;
    ans += mp[a[i]%k];
    mp[a[i]%k]++;
  }
  cout << ans << endl;
  return 0;
}
```