#include <bits/stdc++.h>

using namespace std;
using ll = long long;

signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);

	int n, k;
	cin >> n >> k;

	if (!k) {
		cout << "0\n";
		return 0;
	}

	map<int, int> cnt;
	cnt[1] = 1;

	ll ans = 0;

	vector<ll> p(n + 1, 0);

	for (int i = 0; i < n; i++) {
		ll x;
		cin >> x;
		p[i + 1] = (p[i] + x) % k;

		if (i >= k - 1) {
			cnt[p[i - k + 1] - i + k]--;
		}

		ans += cnt[k + p[i + 1] - i] + cnt[p[i + 1] - i];

		cnt[p[i + 1] - i]++;
	}

	cout << ans;
}
### Prompt

Create a solution in cpp for 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;
using ll = long long;

signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);

	int n, k;
	cin >> n >> k;

	if (!k) {
		cout << "0\n";
		return 0;
	}

	map<int, int> cnt;
	cnt[1] = 1;

	ll ans = 0;

	vector<ll> p(n + 1, 0);

	for (int i = 0; i < n; i++) {
		ll x;
		cin >> x;
		p[i + 1] = (p[i] + x) % k;

		if (i >= k - 1) {
			cnt[p[i - k + 1] - i + k]--;
		}

		ans += cnt[k + p[i + 1] - i] + cnt[p[i + 1] - i];

		cnt[p[i + 1] - i]++;
	}

	cout << ans;
}
```