#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
map<int, int> dp;
int sum[200005];
int main(){
    int n, k, a;
    scanf("%d%d", &n, &k);

    LL ans = 0;
    dp[0] = 1;
    int l = 0;
    for(int i=1;i<=n;i++){
        while(i-l >= k){
            dp[sum[l]]--;
            l++;
        }
        scanf("%d", &a);
        sum[i] = sum[i-1] + a-1;
        sum[i] %= k;
        ans += dp[sum[i]];
        dp[sum[i]]++;
    }
    printf("%lld\n", ans);
}
### 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;
typedef long long LL;
map<int, int> dp;
int sum[200005];
int main(){
    int n, k, a;
    scanf("%d%d", &n, &k);

    LL ans = 0;
    dp[0] = 1;
    int l = 0;
    for(int i=1;i<=n;i++){
        while(i-l >= k){
            dp[sum[l]]--;
            l++;
        }
        scanf("%d", &a);
        sum[i] = sum[i-1] + a-1;
        sum[i] %= k;
        ans += dp[sum[i]];
        dp[sum[i]]++;
    }
    printf("%lld\n", ans);
}
```