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 = [float("inf")] * n
dp[0] = 0
def chmin(a,b):
  if a>b:
    a = b
for i in range(1,n):
  chmin(dp[i],dp[i-1]+abs(h[i]-h[i-1]))
  if i > 1:
    chmin(dp[i],dp[i-2]+abs(h[i]-h[i-2]))
print(dp[-1])
n = int(input())
h = list(map(int,input().split()))
h_min = [0]
h_min.append(abs(h[1]-h[0]))
for i in range(2,n):
  h_min.append(min(abs(h[i]-h[i-1])+h_min[i-1],abs(h[i]-h[i-2])+h_min[i-2]))
print(h_min[-1])