// Author: getharshkumar

#include <bits/stdc++.h>
using namespace std;
#define ll int64_t
const ll N = 1e6 + 6, M = 1e9 + 7;
ll a[N];
int main()
{
    ios::sync_with_stdio(false), cin.tie(nullptr);
    ll n, k, s = 0;
    cin >> n >> k;
    map<ll, ll> m;
    for (ll i = 1; i <= n; i++)
        cin >> a[i], a[i] += a[i - 1];
    m[0] = 1;
    for (ll i = 1, j = 0; i <= n; i++)
    {
        if (i >= k)
            m[(a[j] - j) % k]--, j++;
        s += m[(a[i] - i) % k]++;
    }
    cout << s;
    return 0;
}
### Prompt

Please provide a Cpp coded solution 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
// Author: getharshkumar

#include <bits/stdc++.h>
using namespace std;
#define ll int64_t
const ll N = 1e6 + 6, M = 1e9 + 7;
ll a[N];
int main()
{
    ios::sync_with_stdio(false), cin.tie(nullptr);
    ll n, k, s = 0;
    cin >> n >> k;
    map<ll, ll> m;
    for (ll i = 1; i <= n; i++)
        cin >> a[i], a[i] += a[i - 1];
    m[0] = 1;
    for (ll i = 1, j = 0; i <= n; i++)
    {
        if (i >= k)
            m[(a[j] - j) % k]--, j++;
        s += m[(a[i] - i) % k]++;
    }
    cout << s;
    return 0;
}
```