#include <iostream>
#include <cstdio>
#include <queue>
#include <vector>
#include <map>
#define pb push_back
using namespace std;
int n,k,a[200005];
int64_t ans;
int main(){
    scanf("%d%d",&n,&k);
    for (int i=1;i<=n;++i) scanf("%d",&a[i]),a[i]=(a[i-1]+a[i])%k;
    map<int,int> d;
    ++d[0];
    for (int i=1,j=0;i<=n;++i){
        if (i-j>=k) --d[a[j++]];
        a[i]=(a[i]-i)%k;
        if (a[i]<0) a[i]+=k;
        ans+=d[a[i]];
        ++d[a[i]];
    }
    printf("%lld",ans);
}

### 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 <iostream>
#include <cstdio>
#include <queue>
#include <vector>
#include <map>
#define pb push_back
using namespace std;
int n,k,a[200005];
int64_t ans;
int main(){
    scanf("%d%d",&n,&k);
    for (int i=1;i<=n;++i) scanf("%d",&a[i]),a[i]=(a[i-1]+a[i])%k;
    map<int,int> d;
    ++d[0];
    for (int i=1,j=0;i<=n;++i){
        if (i-j>=k) --d[a[j++]];
        a[i]=(a[i]-i)%k;
        if (a[i]<0) a[i]+=k;
        ans+=d[a[i]];
        ++d[a[i]];
    }
    printf("%lld",ans);
}

```