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 Stone i + 1 or Stone i + 2. 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.

n=int(input())
h=list(map(int,input().split()))
dp=[0 for _ in range(n)]
dp[0]=0
dp[1]=h[0]
for i in range(2,n):
    x=dp[i-1]+abs(h[i-1]-h[i])
    x1 = dp[i - 2] + abs(h[i - 2] - h[i])
    dp[i]=min(x,x1)
print(dp[-1])
n=int(input())
h=list(map(int,input().split()))
dp=[0 for _ in range(n)]
dp[0]=0
dp[1]=abs(h[0]-h[1])
for i in range(2,n):
    x=dp[i-1]+abs(h[i-1]-h[i])
    x1 = dp[i - 2] + abs(h[i - 2] - h[i])
    dp[i]=min(x,x1)
print(dp[-1])