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: Time Limit Exceeded)



import java.util.Scanner;

public class Main {

	static long mincost(int i, long c, long[] ar, long[] dp) {

		if (i == ar.length - 1)
			return 0;

		if (dp[i] != -1)
			return dp[i];

		long mincost = Long.MAX_VALUE;
		
		for(int index = i + 1; index < ar.length; index++)
			mincost = Math.min(mincost, ((Math.abs(ar[i] - ar[index]) * Math.abs(ar[i] - ar[index])) + c + mincost(index, c, ar, dp)));

		return dp[i] = mincost;
	}

	public static void main(String args[]) throws Exception {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		long c = sc.nextLong();
		long ar[] = new long[n];
		long dp[] = new long[n];

		for (int i = 0; i < n; i++) {
			ar[i] = sc.nextLong();
			dp[i] = -1;
		}

		System.out.println(mincost(0, c, ar, dp));
		sc.close();
	}

}