#include <iostream>
#include <cstdio>
#include <map>

using namespace std;

typedef long long ll;

ll n, k;
ll a[200005];
ll b[200005];
map<ll, ll> ma;
ll ans;

int main()
{
    scanf("%lld%lld", &n, &k);
    ma[0]++;
    for(ll i = 1; i <= n; i++)
    {
        scanf("%lld", &a[i]);
        a[i]--;
        b[i] = b[i - 1] + a[i];
        ma[b[i] % k]++;
        if(i >= k) ma[b[i - k] % k]--;
        ans += ma[b[i] % k] - 1;
    }
    printf("%lld\n", ans);
    return 0;
}

### Prompt

In cpp, your task is to solve 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 <iostream>
#include <cstdio>
#include <map>

using namespace std;

typedef long long ll;

ll n, k;
ll a[200005];
ll b[200005];
map<ll, ll> ma;
ll ans;

int main()
{
    scanf("%lld%lld", &n, &k);
    ma[0]++;
    for(ll i = 1; i <= n; i++)
    {
        scanf("%lld", &a[i]);
        a[i]--;
        b[i] = b[i - 1] + a[i];
        ma[b[i] % k]++;
        if(i >= k) ma[b[i - k] % k]--;
        ans += ma[b[i] % k] - 1;
    }
    printf("%lld\n", ans);
    return 0;
}

```