Problem:
Score : 100 points

Problem Statement

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.
Here, h_1 < h_2 < \cdots < h_N holds.

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 one of the following: Stone i + 1, i + 2, \ldots, N. Here, a cost of (h_j - h_i)^2 + C is incurred, where j is the stone to land on.

Find the minimum possible total cost incurred before the frog reaches Stone N.

Constraints

All values in input are integers.

2 \leq N \leq 2 \times 10^5

1 \leq C \leq 10^{12}

1 \leq h_1 < h_2 < \cdots < h_N \leq 10^6

Input

Input is given from Standard Input in the following format:

N C
h_1 h_2 \ldots h_N

Output

Print the minimum possible total cost incurred.

Sample Input 1

5 6
1 2 3 4 5

Sample Output 1

20

If we follow the path 1 → 3 → 5, the total cost incurred would be ((3 - 1)^2 + 6) + ((5 - 3)^2 + 6) = 20.

Sample Input 2

2 1000000000000
500000 1000000

Sample Output 2

1250000000000

The answer may not fit into a 32-bit integer type.

Sample Input 3

8 5
1 3 4 5 10 11 12 13

Sample Output 3

62

If we follow the path 1 → 2 → 4 → 5 → 8, the total cost incurred would be ((3 - 1)^2 + 5) + ((5 - 3)^2 + 5) + ((10 - 5)^2 + 5) + ((13 - 10)^2 + 5) = 62.

Solution in Ruby (2.3.3) (judge verdict: Accepted)

def f(a1,a2,a3,b1,b2,b3)
    return (b3-b2)*(a2-a1) >= (b2-b1)*(a3-a2)
end

n,c = gets.split.map(&:to_i)
h = gets.split.map(&:to_i)
dp = Array.new(n,0)
b = [0]
for i in 1..n-1
    while b.size > 1 && dp[b[0]]+(h[i]-h[b[0]])**2 > dp[b[1]]+(h[i]-h[b[1]])**2
        b.shift
    end
    dp[i] = dp[b[0]]+(h[i]-h[b[0]])**2+c
    while b.size > 1 && f(-2*h[b[-2]],-2*h[b[-1]],-2*h[i],dp[b[-2]]+h[b[-2]]**2,dp[b[-1]]+h[b[-1]]**2,dp[i]+h[i]**2)
        b.pop
    end
    b.push(i)
end
p dp[-1]