#include <bits/stdc++.h>
using namespace std;
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,n)   FOR(i,0,n)
#define pb emplace_back
typedef long long ll;
typedef pair<int,int> pint;

int n,k;
ll a[200001],sum[200001];
map<ll,int> mp;
int main(){
    cin>>n>>k;
    rep(i,n) cin>>a[i];
    rep(i,n) sum[i+1]=(sum[i]+a[i]-1)%k;
    ll ans=0;
    FOR(i,0,n+1){
        if(i>=k) --mp[sum[i-k]];
        ans+=mp[sum[i]];
        ++mp[sum[i]];
    }
    cout<<ans<<endl;
    return 0;
}
### 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 <bits/stdc++.h>
using namespace std;
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,n)   FOR(i,0,n)
#define pb emplace_back
typedef long long ll;
typedef pair<int,int> pint;

int n,k;
ll a[200001],sum[200001];
map<ll,int> mp;
int main(){
    cin>>n>>k;
    rep(i,n) cin>>a[i];
    rep(i,n) sum[i+1]=(sum[i]+a[i]-1)%k;
    ll ans=0;
    FOR(i,0,n+1){
        if(i>=k) --mp[sum[i-k]];
        ans+=mp[sum[i]];
        ++mp[sum[i]];
    }
    cout<<ans<<endl;
    return 0;
}
```