#include <cstdio>
#include <map>
using namespace std;

int pi[233333],pp[233333];
map<int,int> pk;

int main()
{
	int n,k; scanf("%d%d",&n,&k);
	for(int i=1; i<=n; i++) scanf("%d",&pi[i]);
	int p=0; pk[0]++; long long sum=0;
	for(int i=1; i<=n; i++)
	{
		p=(p+pi[i])%k; pp[i]=p;
		if(i-k>=0) pk[(pp[i-k]+k-(i-k)%k)%k]--;
		sum+=pk[(p+k-i%k)%k];
		pk[(p+k-i%k)%k]++;
	}
	printf("%lld\n",sum);
	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 <cstdio>
#include <map>
using namespace std;

int pi[233333],pp[233333];
map<int,int> pk;

int main()
{
	int n,k; scanf("%d%d",&n,&k);
	for(int i=1; i<=n; i++) scanf("%d",&pi[i]);
	int p=0; pk[0]++; long long sum=0;
	for(int i=1; i<=n; i++)
	{
		p=(p+pi[i])%k; pp[i]=p;
		if(i-k>=0) pk[(pp[i-k]+k-(i-k)%k)%k]--;
		sum+=pk[(p+k-i%k)%k];
		pk[(p+k-i%k)%k]++;
	}
	printf("%lld\n",sum);
	return 0;
}
```