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

int main(){
	ll N,K;
	cin >> N >> K;
	vector<ll> A(N);
	for(int i=0;i<N;i++){
		cin >> A[i];
		A[i]--;
	}

	ll ans=0,sum=0;
	map<ll,ll> ma;
	ma[0]++;
	vector<ll> clear;
	for(int i=0;i<N;i++){
		if(i==K-1)ma[0]--;
		if(i-K>=0)ma[clear[i-K]]--;
		(sum += A[i] ) %= K;
		ans += ma[sum];
		ma[sum]++;
		clear.push_back(sum);
	}
	cout << ans << endl;
}
### Prompt

Please create a solution in CPP 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>
#include <chrono>
using namespace std;
typedef long long ll;

int main(){
	ll N,K;
	cin >> N >> K;
	vector<ll> A(N);
	for(int i=0;i<N;i++){
		cin >> A[i];
		A[i]--;
	}

	ll ans=0,sum=0;
	map<ll,ll> ma;
	ma[0]++;
	vector<ll> clear;
	for(int i=0;i<N;i++){
		if(i==K-1)ma[0]--;
		if(i-K>=0)ma[clear[i-K]]--;
		(sum += A[i] ) %= K;
		ans += ma[sum];
		ma[sum]++;
		clear.push_back(sum);
	}
	cout << ans << endl;
}
```