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()))
# N = 4
# h = [10, 30, 40, 20]
dp = [1000000]*N

dp[0]=0
for i in range(1,N):
    if i == 1:
        dp[i] = dp[i-1]+abs(h[i]-h[i-1])
    else:
        dp[i] = min(dp[i-1]+abs(h[i]-h[i-1]), dp[i-2]+abs(h[i]-h[i-2]))
print(dp)
print("min cost=",dp[-1])
N=int(input())
h=list(map(int,input().split()))
# N = 4
# h = [10, 30, 40, 20]
dp = [1000000]*N

dp[0]=0
for i in range(1,N):
    if i == 1:
        dp[i] = dp[i-1]+abs(h[i]-h[i-1])
    else:
        dp[i] = min(dp[i-1]+abs(h[i]-h[i-1]), dp[i-2]+abs(h[i]-h[i-2]))
# print(dp)
# print("min cost=",dp[-1])
print(dp[-1])