#include<cstdio>
#include<cstdlib>
#include<map>
using namespace std;
int m,n,i,j;
#define maxn 200010
long long a[maxn],sum[maxn],k,ans;
map<long long,long long>mp;
int main()
{
	scanf("%d%lld",&n,&k);
	for(i=1;i<=n;i++)
	{
		scanf("%lld",&a[i]);
		sum[i]=sum[i-1]+a[i];
	}
	for(i=n;i>=1;i--)
	{
		mp[(sum[i]-i)%k]++;
		if(n-i+1>=k)
		{
			int p=i+k-1;
			mp[(sum[p]-p)%k]--;
		}
		ans+=mp[(sum[i-1]-i+1)%k];
	}

	printf("%lld\n",ans);
	return 0;
}
### Prompt

Your task is to create 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<cstdlib>
#include<map>
using namespace std;
int m,n,i,j;
#define maxn 200010
long long a[maxn],sum[maxn],k,ans;
map<long long,long long>mp;
int main()
{
	scanf("%d%lld",&n,&k);
	for(i=1;i<=n;i++)
	{
		scanf("%lld",&a[i]);
		sum[i]=sum[i-1]+a[i];
	}
	for(i=n;i>=1;i--)
	{
		mp[(sum[i]-i)%k]++;
		if(n-i+1>=k)
		{
			int p=i+k-1;
			mp[(sum[p]-p)%k]--;
		}
		ans+=mp[(sum[i-1]-i+1)%k];
	}

	printf("%lld\n",ans);
	return 0;
}
```