#include <bits/stdc++.h>
using namespace std;
using ll = long long;

int main(){
	ll N, K;
	cin >> N >> K;
	vector<ll> A(N);
	for(auto &a : A) cin >> a;
	vector<ll> sum(N+1, 0);
	for(int i=1;i<=N;i++) {
		sum[i] = sum[i-1] + A[i-1];
	}
	map<ll,ll> mp;
	ll ans = 0;
	for(int i=0;i<=N;i++) {
		ll plus = (sum[i] - i) % K;
		if(plus < 0) plus += K;
		mp[plus]++;
		if(i >= K) {
			ll minus = (sum[i-K] - (i-K)) % K;
			mp[minus]--;
		}
		// 自分自身との組
		ans += (mp[plus] - 1);
	}
	cout << ans << endl;
}
### 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 <bits/stdc++.h>
using namespace std;
using ll = long long;

int main(){
	ll N, K;
	cin >> N >> K;
	vector<ll> A(N);
	for(auto &a : A) cin >> a;
	vector<ll> sum(N+1, 0);
	for(int i=1;i<=N;i++) {
		sum[i] = sum[i-1] + A[i-1];
	}
	map<ll,ll> mp;
	ll ans = 0;
	for(int i=0;i<=N;i++) {
		ll plus = (sum[i] - i) % K;
		if(plus < 0) plus += K;
		mp[plus]++;
		if(i >= K) {
			ll minus = (sum[i-K] - (i-K)) % K;
			mp[minus]--;
		}
		// 自分自身との組
		ans += (mp[plus] - 1);
	}
	cout << ans << endl;
}
```