#include <bits/stdc++.h>

using namespace std;

#define int long long

int n,k;
map<int,int> cnt;
int a[200001];
int psum[200001];

int32_t main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);

	cin >> n >> k;
	int res = 0;
	cnt[0]++;
	for (int i=1;i<=n;i++)
	{
		cin >> a[i];
		if (i >= k)
		{
			int cur = (psum[i-k] - (i-k))%k;
			cnt[cur]--;
		}
		psum[i] = psum[i-1] + a[i];
		int cur = (psum[i] - i)%k;
		res += cnt[cur];
		cnt[cur]++;
	}	
	cout << res;
}
### 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;

#define int long long

int n,k;
map<int,int> cnt;
int a[200001];
int psum[200001];

int32_t main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);

	cin >> n >> k;
	int res = 0;
	cnt[0]++;
	for (int i=1;i<=n;i++)
	{
		cin >> a[i];
		if (i >= k)
		{
			int cur = (psum[i-k] - (i-k))%k;
			cnt[cur]--;
		}
		psum[i] = psum[i-1] + a[i];
		int cur = (psum[i] - i)%k;
		res += cnt[cur];
		cnt[cur]++;
	}	
	cout << res;
}
```