#include <bits/stdc++.h>
using namespace std;
using ll = long long;

int main()
{
    int N, K;
    cin >> N >> K;
    vector<int> A(N), sum(N + 1);
    for(int i = 0;i < N;i++){
        cin >> A[i]; --A[i];
        sum[i + 1] = (sum[i] + A[i]) % K;
    }
    map<int, int> cnt;
    cnt[0] = 1;
    ll res = 0;
    for(int i = 0;i < N;i++){
        if(K <= i + 1){
            --cnt[sum[i + 1 - K]];
        }
        res += cnt[sum[i + 1]];
        ++cnt[sum[i + 1]];
    }
    cout << res << endl;
    return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
using ll = long long;

int main()
{
    int N, K;
    cin >> N >> K;
    vector<int> A(N), sum(N + 1);
    for(int i = 0;i < N;i++){
        cin >> A[i]; --A[i];
        sum[i + 1] = (sum[i] + A[i]) % K;
    }
    map<int, int> cnt;
    cnt[0] = 1;
    ll res = 0;
    for(int i = 0;i < N;i++){
        if(K <= i + 1){
            --cnt[sum[i + 1 - K]];
        }
        res += cnt[sum[i + 1]];
        ++cnt[sum[i + 1]];
    }
    cout << res << endl;
    return 0;
}

```