#include <bits/stdc++.h>

int ri() {
    int n;
    scanf("%d", &n);
    return n;
}
int main() {
	int n = ri(), k = ri();
	int sum[n + 1];
	sum[0] = 0;
	std::map<int, int> ma{{0, 1}};
	int64_t res = 0;
	for (int i = 1; i <= n; i++) {
		if (i >= k) ma[sum[i - k]]--;
		int64_t a = ri() - 1 + k;
		sum[i] = (sum[i - 1] + a) % k;
		res += ma[sum[i]];
		ma[sum[i]]++;
	}
	std::cout << res << std::endl;
    return 0;
}

### Prompt

In cpp, your task is to solve 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>

int ri() {
    int n;
    scanf("%d", &n);
    return n;
}
int main() {
	int n = ri(), k = ri();
	int sum[n + 1];
	sum[0] = 0;
	std::map<int, int> ma{{0, 1}};
	int64_t res = 0;
	for (int i = 1; i <= n; i++) {
		if (i >= k) ma[sum[i - k]]--;
		int64_t a = ri() - 1 + k;
		sum[i] = (sum[i - 1] + a) % k;
		res += ma[sum[i]];
		ma[sum[i]]++;
	}
	std::cout << res << std::endl;
    return 0;
}

```