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

int n, k;
int S[200005];
int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> n >> k;
    for (int i = 1; i <= n; i++) {
        cin >> S[i];
        S[i] += S[i - 1] - 1;
        S[i] %= k;
    }
    map<int, int> cnt;
    ll ans = 0;
    for (int i = 1; i <= n; i++) {
        if (i - k >= 0) cnt[S[i - k]]--;
        cnt[S[i - 1]]++;
        ans += cnt[S[i]];
    }
    cout << ans;
}
### 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;
typedef long long ll;

int n, k;
int S[200005];
int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> n >> k;
    for (int i = 1; i <= n; i++) {
        cin >> S[i];
        S[i] += S[i - 1] - 1;
        S[i] %= k;
    }
    map<int, int> cnt;
    ll ans = 0;
    for (int i = 1; i <= n; i++) {
        if (i - k >= 0) cnt[S[i - k]]--;
        cnt[S[i - 1]]++;
        ans += cnt[S[i]];
    }
    cout << ans;
}
```