#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int MAXN=1000010;
int n,k,a,s[MAXN];
ll ans;
map <int,int> m;
int main () {
	scanf("%d%d",&n,&k);
	m[0]++;
	for (int i=1;i<=n;i++) {
		scanf("%d",&a);
		s[i+1]=((s[i]+a-1)%k+k)%k;
		if (i-k+1>0) {m[s[i-k+1]]--;}
		ans+=m[s[i+1]];
		m[s[i+1]]++;
	}
	printf("%lld\n",ans);
	return 0;
}
### Prompt

Develop a solution in CPP to the problem described below:

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 ll long long
using namespace std;
const int MAXN=1000010;
int n,k,a,s[MAXN];
ll ans;
map <int,int> m;
int main () {
	scanf("%d%d",&n,&k);
	m[0]++;
	for (int i=1;i<=n;i++) {
		scanf("%d",&a);
		s[i+1]=((s[i]+a-1)%k+k)%k;
		if (i-k+1>0) {m[s[i-k+1]]--;}
		ans+=m[s[i+1]];
		m[s[i+1]]++;
	}
	printf("%lld\n",ans);
	return 0;
}
```