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 = [int(s) for s in input().split()]

dp = [0] * n
dp[0],dp[1] = 0,abs(dp[1]-dp[0])

for i in range(2,n):
  dp[i] = min(dp[i-1] + abs(h[i]-h[i-1]),
              dp[i-2] + abs(h[i]-h[i-2]))
 
print(dp[-1])
N = int(input())
h = [int(s) for s in input().split()]
dp = [0]*N
for i in range(N):
    if i == 0:
        dp[0] = 0
    elif i == 1:
        dp[i] = abs(h[i]-h[i-1])+dp[i-1]
    else:
        dp[i] = min(abs(h[i]-h[i-1])+dp[i-1], abs(h[i]-h[i-2])+dp[i-2])

print(dp[N-1])
