#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(){
    int n,k;
    scanf("%d%d",&n,&k);
    int a[n+1];
    unordered_map<int,int> ct;
    ll ans = 0;
    int p[n+1];
    p[0] = 0;
    ct[0] = 1;
    for (int i = 1; i <= n; i++){
        scanf("%d",&a[i]);
        a[i]--;
        p[i] = p[i-1] + a[i];
        p[i] %= k;
        if (i-k >= 0) ct[p[i-k]]--;
        //printf("%d %d\n",p[i],ct[p[i]]);
        ans += ct[p[i]];
        ct[p[i]]++;

    }
    printf("%lld",ans);

}

### 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>
using namespace std;
typedef long long ll;
int main(){
    int n,k;
    scanf("%d%d",&n,&k);
    int a[n+1];
    unordered_map<int,int> ct;
    ll ans = 0;
    int p[n+1];
    p[0] = 0;
    ct[0] = 1;
    for (int i = 1; i <= n; i++){
        scanf("%d",&a[i]);
        a[i]--;
        p[i] = p[i-1] + a[i];
        p[i] %= k;
        if (i-k >= 0) ct[p[i-k]]--;
        //printf("%d %d\n",p[i],ct[p[i]]);
        ans += ct[p[i]];
        ct[p[i]]++;

    }
    printf("%lld",ans);

}

```