#include <iostream>
#include <map>
#include <vector>
#include <numeric>

using namespace std;

int main(){
    int n, k;
    cin >> n >> k;
    vector<long long> a(n), b(n + 1);
    for(auto&& i : a){
        cin >> i;
    }
    map<long long, long long> cnt;
    cnt[0] = 1;
    long long ans = 0;
    for(int i = 1; i <= n; i++){
        b[i] = b[i - 1] + a[i - 1] + k - 1;
        if(i - k >= 0) cnt[b[i - k] % k]--;
        ans += cnt[b[i] % k];
        cnt[b[i]%k]++;
    }
    cout << ans << "\n";
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 <map>
#include <vector>
#include <numeric>

using namespace std;

int main(){
    int n, k;
    cin >> n >> k;
    vector<long long> a(n), b(n + 1);
    for(auto&& i : a){
        cin >> i;
    }
    map<long long, long long> cnt;
    cnt[0] = 1;
    long long ans = 0;
    for(int i = 1; i <= n; i++){
        b[i] = b[i - 1] + a[i - 1] + k - 1;
        if(i - k >= 0) cnt[b[i - k] % k]--;
        ans += cnt[b[i] % k];
        cnt[b[i]%k]++;
    }
    cout << ans << "\n";
}

```