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()))
diff = 0
prev = l[0]
i = 1
while(i < n-1):
    diff += prev-min(l[i],l[i+1])
    prev = min(l[i],l[i+1])
    if(prev == l[i]):
        i += 1
    else:
        i += 2
print(diff)

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