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

using i32 = int_fast32_t;
using i64 = int_fast64_t;

int main() {
    i32 n, k;
    cin >> n >> k;
    vector<i64> a(n+1);
    for (i32 i = 1; i < n+1; i++) cin >> a[i];
    partial_sum(begin(a), end(a), begin(a));

    i32 ans = 0;
    map<i32, i32> count;
    for (i32 i = 0; i < n+1; i++) {
        if (i >= k) count[(a[i-k] - (i-k)) % k]--;
        ans += count[(a[i] - i) % k]++;
    }

    cout << ans << endl;
}

### Prompt

Your challenge is to write 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;

using i32 = int_fast32_t;
using i64 = int_fast64_t;

int main() {
    i32 n, k;
    cin >> n >> k;
    vector<i64> a(n+1);
    for (i32 i = 1; i < n+1; i++) cin >> a[i];
    partial_sum(begin(a), end(a), begin(a));

    i32 ans = 0;
    map<i32, i32> count;
    for (i32 i = 0; i < n+1; i++) {
        if (i >= k) count[(a[i-k] - (i-k)) % k]--;
        ans += count[(a[i] - i) % k]++;
    }

    cout << ans << endl;
}

```