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 C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Linq;
using System.Collections;
using System.Collections.Generic;

class Program
{
    public static void Main()
    {
        string[] s = Console.ReadLine().Split(' ');
        int N = int.Parse(s[0]);
        long C = long.Parse(s[1]);
        long[] h = new long[N];
        s = Console.ReadLine().Split(' ');
        for (int i = 0; i < N; i++)
            h[i] = int.Parse(s[i]);

        long[] dp = new long[N];
        for (int i = 0; i < N; i++)
            dp[i] = long.MaxValue;

        dp[0] = 0;
        for (int i = 1; i < N; i++)
        {
            for (int j = i - 1; j >= 0 && dp[j] + (h[i] - h[j]) * (h[i] - h[j]) + C <= dp[i]; j--)
            {
                dp[i] = dp[j] + (h[i] - h[j]) * (h[i] - h[j]) + C;
            }
        }
        Console.WriteLine(dp[N - 1]);
    }
}