#include <bits/stdc++.h>
using namespace std;
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int n, k;
	cin >> n >> k;
	vector<int> a;
	a.resize(n + 1);
	for(int i = 1; i <= n; ++i) {
		cin >> a[i];
	}
	map<int, int> mp;
	long long ans = 0;
	++mp[0];
	for(int i = 1; i <= n; ++i) {
		a[i] = (a[i] + a[i - 1]) % k;
		int t = ((a[i] - i) % k + k) % k;
		if(i - k >= 0) {
			--mp[((a[i - k] - (i - k)) % k + k) % k];
		}
		ans += mp[t];
		++mp[t];
	}
	cout << ans << '\n';
	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>
using namespace std;
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int n, k;
	cin >> n >> k;
	vector<int> a;
	a.resize(n + 1);
	for(int i = 1; i <= n; ++i) {
		cin >> a[i];
	}
	map<int, int> mp;
	long long ans = 0;
	++mp[0];
	for(int i = 1; i <= n; ++i) {
		a[i] = (a[i] + a[i - 1]) % k;
		int t = ((a[i] - i) % k + k) % k;
		if(i - k >= 0) {
			--mp[((a[i - k] - (i - k)) % k + k) % k];
		}
		ans += mp[t];
		++mp[t];
	}
	cout << ans << '\n';
	return 0;
}

```