#include<bits/stdc++.h>

using namespace std;
using ll = long long;

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int n, k;
    cin >> n >> k;
    ll ans = 0;
    map<ll, int> cnt;
    cnt[0] = 1;
    vector<ll> p(n + 1);
    for (int i = 1; i <= n; ++i) {
        if (i >= k)
            cnt[p[i - k]]--;
        cin >> p[i];
        p[i] = (p[i] - 1) % k;
        p[i] = (p[i] + p[i - 1]) % k;
        ans += cnt[p[i]]++;
    }
    cout << ans;
    return 0;
}
### Prompt

Develop a solution in CPP 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<bits/stdc++.h>

using namespace std;
using ll = long long;

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