#include<iostream>
#include<map>
using namespace std;
long long a[200005],s[200005];
map<long long,long long> mod;
int main()
{
	long long n,k,sum=0;
	scanf("%lld%lld",&n,&k);
	for(int i=1;i<=n;i++)scanf("%lld",&a[i]);
	for(int i=1;i<=n;i++)s[i]=s[i-1]+a[i];
	s[0]=0;a[0]=0;
	for(int i=1;i<=n;i++)
	{
		mod[(s[i-1]-i+1)%k]++;
		if(i>=k)mod[(s[i-k]-(i-k))%k]--;
		sum+=mod[(s[i]-i)%k];
	}
	printf("%lld\n",sum);
}
### 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<iostream>
#include<map>
using namespace std;
long long a[200005],s[200005];
map<long long,long long> mod;
int main()
{
	long long n,k,sum=0;
	scanf("%lld%lld",&n,&k);
	for(int i=1;i<=n;i++)scanf("%lld",&a[i]);
	for(int i=1;i<=n;i++)s[i]=s[i-1]+a[i];
	s[0]=0;a[0]=0;
	for(int i=1;i<=n;i++)
	{
		mod[(s[i-1]-i+1)%k]++;
		if(i>=k)mod[(s[i-k]-(i-k))%k]--;
		sum+=mod[(s[i]-i)%k];
	}
	printf("%lld\n",sum);
}
```