#define HAVE_STRUCT_TIMESPEC
#include<bits/stdc++.h>
using namespace std;
long long a[200007];
map<long long,long long>b;
int main(){
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long n,k;
    cin>>n>>k;
    for(int i=1;i<=n;++i)
        cin>>a[i];
    for(int i=1;i<=n;++i)
        a[i]=(a[i-1]+a[i]-1)%k;
    long long ans=0;
    b[0]=1;
    for(int l=0,r=1;r<=n;++r){
        if(r-l==k)
            --b[a[l++]];
        ans+=b[a[r]];
        ++b[a[r]];
    }
    cout<<ans;
    return 0;
}

### Prompt

Your challenge is to write 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
#define HAVE_STRUCT_TIMESPEC
#include<bits/stdc++.h>
using namespace std;
long long a[200007];
map<long long,long long>b;
int main(){
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long n,k;
    cin>>n>>k;
    for(int i=1;i<=n;++i)
        cin>>a[i];
    for(int i=1;i<=n;++i)
        a[i]=(a[i-1]+a[i]-1)%k;
    long long ans=0;
    b[0]=1;
    for(int l=0,r=1;r<=n;++r){
        if(r-l==k)
            --b[a[l++]];
        ans+=b[a[r]];
        ++b[a[r]];
    }
    cout<<ans;
    return 0;
}

```