#include <stdio.h>
#include <algorithm>
int n, k, a;
long long e, x, b, i, c[200001];
int main() {
	scanf("%d%d", &n, &k);
	for (i = 1; i <= n; i++) {
		scanf("%d", &a);
		b = (a + b - 1) % k;
		c[i] = b * (k + n) + i;
	}
	std::sort(c, c + n + 1);
	for (i = 1; i <= n; i++) {
		while (c[i] - c[x] >= k)x++;
		e += i - x;
	}
	printf("%lld\n", e);
}
### Prompt

Please formulate 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 <stdio.h>
#include <algorithm>
int n, k, a;
long long e, x, b, i, c[200001];
int main() {
	scanf("%d%d", &n, &k);
	for (i = 1; i <= n; i++) {
		scanf("%d", &a);
		b = (a + b - 1) % k;
		c[i] = b * (k + n) + i;
	}
	std::sort(c, c + n + 1);
	for (i = 1; i <= n; i++) {
		while (c[i] - c[x] >= k)x++;
		e += i - x;
	}
	printf("%lld\n", e);
}
```