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 Java8 (OpenJDK 1.8.0) (judge verdict: Runtime Error)

import java.util.*;

public class Main {

  public static void main(String[] args) {
    Scanner scan = new Scanner(System.in);
    String[] hiki = scan.nextLine().split(" ");
    int n = Integer.parseInt(hiki[0]);
    int C = Integer.parseInt(hiki[1]);
    long[] cost = new long[n];
    long[] dp = new long[n];
    hiki = scan.nextLine().split(" ");
    for(int i = 0;i < n;i++){
      cost[i] = Integer.parseInt(hiki[i]);
      dp[i]=Long.MAX_VALUE;
    }
    dp[0]=0;
    for(int i = 1;i < n;i++){
      for(int j = 1;j < i+1;j ++){
        dp[i] = Math.min(dp[i],
        dp[i-j]+(cost[i]-cost[i-j])*
        (cost[i]-cost[i-j])+C);
      }
    }
    System.out.println(dp[n-1]);
  }
}