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())
l = list(map(int, input().split()))
l1 = [0]*n
for i in range(1,n):
  l1[i] = min(l[i], abs(l[i]-l[i-1]))
print(l)
n = int(input())
l = list(map(int, input().split()))
l1 = [0]*n
l1[1] = abs(l[0] - l[1])
for i in range(2, n):
    l1[i] = min(l1[i - 2] + abs(l[i] - l[i - 2]), l1[i - 1] + abs(l[i] - l[i - 1]))
print(l1[n - 1])