p03161 Educational DP Contest - Frog 2
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i.

There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N:

* If the frog is currently on Stone i, jump to one of the following: Stone i + 1, i + 2, \ldots, i + K. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on.



Find the minimum possible total cost incurred before the frog reaches Stone N.

Constraints

* All values in input are integers.
* 2 \leq N \leq 10^5
* 1 \leq K \leq 100
* 1 \leq h_i \leq 10^4

Input

Input is given from Standard Input in the following format:


N K
h_1 h_2 \ldots h_N


Output

Print the minimum possible total cost incurred.

Examples

Input

5 3
10 30 40 50 20


Output

30


Input

3 1
10 20 10


Output

20


Input

2 100
10 10


Output

0


Input

10 4
40 10 20 70 80 10 20 70 80 60


Output

40
```python
N,K=map(int,input().split())
h=list(map(int,input().split()))
dp=[10**10]*(N)#dp[i]→i+1番目の柱にくる最小コスト
dp[0]=0
for i in range(N-1):
    dp[i+1]=min(dp[i+1-j]+abs(h[i+1]-h[i+1-j]) for j in range(1,min(i+1,K)+1))
print(dp[N-1])
```