#include<bits/stdc++.h>
using namespace std;
using ll=long long;
int main(){
    ll n;
    ll k;
    cin>>n>>k;
    vector<ll>a(n);
    for(int i=0;i<n;++i){
        cin>>a[i];
    }
    vector<ll>s(n+1,0);
    for(int i=1;i<n+1;++i){
        s[i]=(s[i-1]+a[i-1]-1)%k;
    }
    map<ll,ll>mp;
    for(int i=0;i<min(n+1,k);++i){
        mp[s[i]]++;
    }
    ll ans=0;
    for(int i=0;i<n;++i){
        ans+=--mp[s[i]];
        if(i+k<=n){
            mp[s[i+k]]++;
        }
    }
    cout<<ans<<endl;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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;
    ll k;
    cin>>n>>k;
    vector<ll>a(n);
    for(int i=0;i<n;++i){
        cin>>a[i];
    }
    vector<ll>s(n+1,0);
    for(int i=1;i<n+1;++i){
        s[i]=(s[i-1]+a[i-1]-1)%k;
    }
    map<ll,ll>mp;
    for(int i=0;i<min(n+1,k);++i){
        mp[s[i]]++;
    }
    ll ans=0;
    for(int i=0;i<n;++i){
        ans+=--mp[s[i]];
        if(i+k<=n){
            mp[s[i+k]]++;
        }
    }
    cout<<ans<<endl;
}
```