#include <bits/stdc++.h>

using namespace std;
using ll = long long;

const int N = 200005;

int n, k, a[N];
int p[N];

int main() {

	scanf("%d %d", &n, &k);

	for(int i = 1; i <= n; i++) scanf("%d", a + i);

	map<int, int> cnt;

	int s = 0;

	cnt[s]++;

	ll ans = 0;
	for(int i = 1; i <= n; i++) {
		if(i >= k) {
			cnt[p[i - k] - (i - k)]--;
		}
		s = (s + a[i]) % k;
		p[i] = s;
		ans += cnt[s - i];
		ans += cnt[s - i + k];
		cnt[s - i]++;
	}

	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;
using ll = long long;

const int N = 200005;

int n, k, a[N];
int p[N];

int main() {

	scanf("%d %d", &n, &k);

	for(int i = 1; i <= n; i++) scanf("%d", a + i);

	map<int, int> cnt;

	int s = 0;

	cnt[s]++;

	ll ans = 0;
	for(int i = 1; i <= n; i++) {
		if(i >= k) {
			cnt[p[i - k] - (i - k)]--;
		}
		s = (s + a[i]) % k;
		p[i] = s;
		ans += cnt[s - i];
		ans += cnt[s - i + k];
		cnt[s - i]++;
	}

	printf("%lld\n", ans);
}

```