#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 2e5 + 10;
int n, k;
ll ans;
map<int, int> M;
ll norm(ll x) { return (x % k + k) % k;}
ll a[maxn];
int main()
{
	scanf("%d%d", &n, &k);
	for(int i = 1; i <= n; ++i) scanf("%lld", &a[i]), a[i] += a[i - 1];
	M[0] = 1;
	for(int i = 1; i <= n; ++i) 
	{
		a[i] = norm(a[i] - i);
		if(i >= k) --M[a[i - k]]; 
		ans += M[a[i]];
		++M[a[i]];
	}
	printf("%lld\n", ans);
	return 0;
}
### Prompt

Generate 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;
typedef long long ll;
const int maxn = 2e5 + 10;
int n, k;
ll ans;
map<int, int> M;
ll norm(ll x) { return (x % k + k) % k;}
ll a[maxn];
int main()
{
	scanf("%d%d", &n, &k);
	for(int i = 1; i <= n; ++i) scanf("%lld", &a[i]), a[i] += a[i - 1];
	M[0] = 1;
	for(int i = 1; i <= n; ++i) 
	{
		a[i] = norm(a[i] - i);
		if(i >= k) --M[a[i - k]]; 
		ans += M[a[i]];
		++M[a[i]];
	}
	printf("%lld\n", ans);
	return 0;
}
```