#include<bits/stdc++.h>

using namespace std;

int main(void) {
    long n, k;
    cin >> n >> k;

    vector<long> a(n);
    for(auto &x: a) cin >> x, x--;

    vector<long> b(n+1);
    for(int i=1; i<=n; i++)
        b[i] = (b[i-1] + a[i-1]) % k;

    map<long,long> m;
    long r = 0;
    for(int i=0; i<=n; i++) {
        r += m[b[i]], m[b[i]]++;
        if(i>=k-1)
            m[b[i-(k-1)]]--;
    }

    cout << r << endl;
}

### Prompt

Please create a solution in Cpp 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;

int main(void) {
    long n, k;
    cin >> n >> k;

    vector<long> a(n);
    for(auto &x: a) cin >> x, x--;

    vector<long> b(n+1);
    for(int i=1; i<=n; i++)
        b[i] = (b[i-1] + a[i-1]) % k;

    map<long,long> m;
    long r = 0;
    for(int i=0; i<=n; i++) {
        r += m[b[i]], m[b[i]]++;
        if(i>=k-1)
            m[b[i-(k-1)]]--;
    }

    cout << r << endl;
}

```