#include<bits/stdc++.h>
using namespace std;
#define ll long long
ll a[1000000];
map<ll,ll> num;
ll add[1000000];
int main()
{
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    int n,k;
    cin>>n>>k;
    num[0]=1;
    ll ret=0;
    for (int i=1;i<=n;i++) {
        if (i>=k)num[add[i-k]]--;
        cin>>a[i];
        a[i]+=a[i-1];
        ll have = ((a[i]-i)%k+k)%k;
        add[i] = have;
        ret+=num[have];
        num[((a[i]-i)%k+k)%k]++;
    }
    cout<<ret<<endl;
}

### Prompt

Generate 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
#include<bits/stdc++.h>
using namespace std;
#define ll long long
ll a[1000000];
map<ll,ll> num;
ll add[1000000];
int main()
{
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    int n,k;
    cin>>n>>k;
    num[0]=1;
    ll ret=0;
    for (int i=1;i<=n;i++) {
        if (i>=k)num[add[i-k]]--;
        cin>>a[i];
        a[i]+=a[i-1];
        ll have = ((a[i]-i)%k+k)%k;
        add[i] = have;
        ret+=num[have];
        num[((a[i]-i)%k+k)%k]++;
    }
    cout<<ret<<endl;
}

```