#include<bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0; i<(n); i++)
int n,k,a;
long long b,ans;
map<int,vector<int>> m;
int main(){
	scanf("%d%d", &n, &k);
	m[0].push_back(0);
	rep(i,n){
		scanf("%d", &a);
		b=(b+a-1)%k;
		ans += m[b].end() - upper_bound(m[b].begin(), m[b].end(),i+1-k);
		m[b].push_back(i+1);
	}
	printf("%lld\n", ans);
}
### Prompt

Please provide a CPP coded solution 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>
using namespace std;
#define rep(i,n) for(int i=0; i<(n); i++)
int n,k,a;
long long b,ans;
map<int,vector<int>> m;
int main(){
	scanf("%d%d", &n, &k);
	m[0].push_back(0);
	rep(i,n){
		scanf("%d", &a);
		b=(b+a-1)%k;
		ans += m[b].end() - upper_bound(m[b].begin(), m[b].end(),i+1-k);
		m[b].push_back(i+1);
	}
	printf("%lld\n", ans);
}
```