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

const int N=2e5+20;

int n,k,x;
map <int,queue <int>> m;
long long s,ans;

int main()
{
	scanf("%d%d",&n,&k);
	m[0].push(0);
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&x);
		s+=x;

		int r=(s-i)%k;
		m[r].push(i);

		while(i-m[r].front()>=k) m[r].pop();
		ans+=m[r].size()-1;
	}
	printf("%lld",ans);
}
### Prompt

Construct a cpp code solution to the problem outlined:

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;

const int N=2e5+20;

int n,k,x;
map <int,queue <int>> m;
long long s,ans;

int main()
{
	scanf("%d%d",&n,&k);
	m[0].push(0);
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&x);
		s+=x;

		int r=(s-i)%k;
		m[r].push(i);

		while(i-m[r].front()>=k) m[r].pop();
		ans+=m[r].size()-1;
	}
	printf("%lld",ans);
}
```