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

typedef long long ll;

int n,k,a,f;
map<int,int> mp,pr;
ll ans;

int main(){

    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    cin>>n>>k;
    mp[0]=1;
    for(int i=1;i<=n;i++){
        cin>>a;f=(f+a-1)%k;
        pr[i+1]=f;
        if(i>=k)mp[pr[i-k+1]]--;
        ans+=mp[f]++;
    }
    cout<<ans<<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>
using namespace std;

typedef long long ll;

int n,k,a,f;
map<int,int> mp,pr;
ll ans;

int main(){

    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    cin>>n>>k;
    mp[0]=1;
    for(int i=1;i<=n;i++){
        cin>>a;f=(f+a-1)%k;
        pr[i+1]=f;
        if(i>=k)mp[pr[i-k+1]]--;
        ans+=mp[f]++;
    }
    cout<<ans<<endl;

    return 0;
}
```