#include<bits/stdc++.h>
using namespace std;
int a[200100];
long long sum[200100];
map<long long,int> mp;
int front[200100];
int main(){
	int n,k;
	scanf("%d%d",&n,&k);
	long long ans=0;
	for(int i=1;i<=n;i++)scanf("%d",&a[i]),a[i]%=k;
	for(int i=1;i<=n;i++)sum[i]=sum[i-1]+a[i];
	mp[0]=1;
	int l=1,r=0;
	for(long long i=1;i<=n;i++){
		if(i>=k)mp[(sum[i-k]-(i-k)+k)%k]--;
		if(mp[(sum[i]-i+k)%k]){
			ans+=mp[(sum[i]-i+k)%k];	
		}
		mp[(sum[i]-i+k)%k]++;
	}
	cout<<ans;
		
	return 0;
}
### Prompt

Your challenge is to write 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>
using namespace std;
int a[200100];
long long sum[200100];
map<long long,int> mp;
int front[200100];
int main(){
	int n,k;
	scanf("%d%d",&n,&k);
	long long ans=0;
	for(int i=1;i<=n;i++)scanf("%d",&a[i]),a[i]%=k;
	for(int i=1;i<=n;i++)sum[i]=sum[i-1]+a[i];
	mp[0]=1;
	int l=1,r=0;
	for(long long i=1;i<=n;i++){
		if(i>=k)mp[(sum[i-k]-(i-k)+k)%k]--;
		if(mp[(sum[i]-i+k)%k]){
			ans+=mp[(sum[i]-i+k)%k];	
		}
		mp[(sum[i]-i+k)%k]++;
	}
	cout<<ans;
		
	return 0;
}
```