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 Java (OpenJDK 11.0.6) (judge verdict: Time Limit Exceeded)

import java.util.*;
class Main
{
    // static double total =0.0;
    // public static void select(double arr[],boolean visited[] , int start,double tot,int n, int mid)
    // {
    //     if(start>=n)
    //     {
    //         total +=tot;
    //         System.out.println(tot);
    //         return;
    //     }
    //     for(int i=start;i<n;i++)
    //     {
    //        if(mid>0)
    //         select(arr,visited,i+1,tot*arr[i],n,mid-1);
    //         select(arr,visited,i+1,tot*(1-arr[i]),n,mid);
    //     }
    //     return;
    // }
    // double arr[] = new double[n];
    //     for(int i=0;i<n;i++)
    //     {
    //         arr[i]=sc.nextDouble();
    //     }
    //     boolean visited[]=new boolean[n];
    //     Arrays.fill(visited,false);
    //     int mid = n/2+1;
    //     for(int j=mid;j<=n;j++)
    //     {
    //         double tot = 1;
    //         int start=0;
    //         select(arr,visited,start,tot,n,j);
    //     }
    //     System.out.println(total);
    public static void main(String [] args)
    {
        Scanner sc = new Scanner(System.in);
        int n =sc.nextInt();
        long c = sc.nextLong();
        int arr[]= new int[n];
        long res[]=new long[n];
        for(int i=0;i<n;i++)
        {
            arr[i]=sc.nextInt();
            res[i]=Long.MAX_VALUE;
        }
        res[0]=0;
        for(int i=1;i<n;i++)
        {
            for(int j=0;j<i;j++)
            {
                res[i]= Math.min(res[i],(long)(Math.pow(arr[i]-arr[j],2)+c)+res[j]);
            }
        }
        System.out.println(res[n-1]);
    }
}