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
import sys
# import numpy as np
inp = sys.stdin.readline
read = lambda: list(map(int, inp().strip().split()))
# import profile

def solve():
    n, k =read(); arr = read()
    dp = [-1]*n; dp[0] = 0
    for i in range(n):
    	for j in range(1, k+1):
    		if i-j >-1:
    			tem = dp[i-j]+abs(arr[i-j] - arr[i])
    			if dp[i] == -1:
    				dp[i] = tem
    			elif dp[i] > tem:
    				dp[i] = tem
    				# dp[i] = min(dp[i], tem)
    print(dp[-1])

def one():
	l = [0]*10**7
def two():
	l = [0 for i in range(10**7)]

if __name__ == "__main__":
	solve()
	# cProfile.run(solve())
	# profile.run(one())
	# cProfile.run()