#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int n, k;
    cin >> n >> k;
    vector<int> a(n);
    for(int i=0; i<n; i++)
        cin >> a[i];
    ll dp[n+1];
    dp[0] = 0;
    for(int i=0; i<n; i++)
        dp[i+1] = (dp[i] + a[i] -1) % k;
    map<int, int> mp;
    ll ans = 0;
    mp[0] = 1;
    for(int i=1; i<=n; i++){
        if(i>=k) mp[dp[i-k]]--;
        ans += mp[dp[i]]++;
    }
    cout << ans << endl;
    return 0;
}

### 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 main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int n, k;
    cin >> n >> k;
    vector<int> a(n);
    for(int i=0; i<n; i++)
        cin >> a[i];
    ll dp[n+1];
    dp[0] = 0;
    for(int i=0; i<n; i++)
        dp[i+1] = (dp[i] + a[i] -1) % k;
    map<int, int> mp;
    ll ans = 0;
    mp[0] = 1;
    for(int i=1; i<=n; i++){
        if(i>=k) mp[dp[i-k]]--;
        ans += mp[dp[i]]++;
    }
    cout << ans << endl;
    return 0;
}

```