#include <bits/stdc++.h>

#define rep(i,n) for(int i=0;i<(n);i++)

using namespace std;
using lint=long long;

int main(){
	int n,k; scanf("%d%d",&n,&k);
	vector<lint> a(n);
	rep(i,n) scanf("%lld",&a[i]), a[i]--;

	vector<lint> sum(n+1);
	rep(i,n) sum[i+1]=(sum[i]+a[i])%k;

	map<lint,int> f;
	rep(i,min(n+1,k)) ++f[sum[i]];

	lint ans=0;
	rep(i,n){
		--f[sum[i]];
		ans+=f[sum[i]];
		if(k+i<=n) ++f[sum[k+i]];
	}
	printf("%lld\n",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>

#define rep(i,n) for(int i=0;i<(n);i++)

using namespace std;
using lint=long long;

int main(){
	int n,k; scanf("%d%d",&n,&k);
	vector<lint> a(n);
	rep(i,n) scanf("%lld",&a[i]), a[i]--;

	vector<lint> sum(n+1);
	rep(i,n) sum[i+1]=(sum[i]+a[i])%k;

	map<lint,int> f;
	rep(i,min(n+1,k)) ++f[sum[i]];

	lint ans=0;
	rep(i,n){
		--f[sum[i]];
		ans+=f[sum[i]];
		if(k+i<=n) ++f[sum[k+i]];
	}
	printf("%lld\n",ans);

	return 0;
}

```