#include <bits/stdc++.h>

using namespace std;

long long pref[int(2e5 + 544)];

int main () { ios_base::sync_with_stdio(0); cin.tie(0), cout.tie(0);
	int n, k;
	long long a, p = 0, ans = 0;
	map <long long, long long> mp;
	mp[0] = 1;
	cin >> n >> k;
	for (int i = 1 ; i <= n ; ++ i) {
		cin >> a;
		p += a - 1;
		ans += mp[p % k];
		if (i - k + 1 >= 0) {
			mp[pref[i - k + 1] % k]--;
		}
		mp[p % k]++;
		pref[i] = p;
	}
	cout << ans * (k != 1);
}
### Prompt

Please create a solution in CPP 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;

long long pref[int(2e5 + 544)];

int main () { ios_base::sync_with_stdio(0); cin.tie(0), cout.tie(0);
	int n, k;
	long long a, p = 0, ans = 0;
	map <long long, long long> mp;
	mp[0] = 1;
	cin >> n >> k;
	for (int i = 1 ; i <= n ; ++ i) {
		cin >> a;
		p += a - 1;
		ans += mp[p % k];
		if (i - k + 1 >= 0) {
			mp[pref[i - k + 1] % k]--;
		}
		mp[p % k]++;
		pref[i] = p;
	}
	cout << ans * (k != 1);
}
```