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: Accepted)

import java.util.LinkedList;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        long c = in.nextLong();
        LinkedList<Stone> queue = new LinkedList<>();
        long h1 = in.nextLong();
        queue.add(new Stone(h1, h1 * h1));
        for (int i = 1; i < n; i++) {
            long h = in.nextLong();
            while (queue.size() > 1 && queue.get(1).dp - (2L * h * queue.get(1).h) <= queue.get(0).dp - (2L * h * queue.get(0).h)) {
                queue.removeFirst();
            }
            long dp = queue.get(0).dp - (2L * h * queue.get(0).h) + (2L * h * h) + c;
            Stone stone = new Stone(h, dp);
            //while (!queue.isEmpty() && dp - (2L * h * h) <= queue.getLast().dp - (2L * h * queue.getLast().h)) {
            while (queue.size() > 1 && fromLine(queue.get(queue.size() - 2), queue.getLast(), stone) <= 0L) {
                queue.removeLast();
            }
            queue.addLast(stone);
        }
        System.out.println(queue.getLast().dp - (queue.getLast().h * queue.getLast().h));
    }

    static class Stone {
        final long h;
        final long dp;

        Stone(long h, long dp) {
            this.h = h;
            this.dp = dp;
        }
    }

    static long fromLine(Stone a, Stone b, Stone point) {
        return ((point.dp - a.dp) * (b.h - a.h)) - ((point.h - a.h) * (b.dp - a.dp));
    }
}