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

public class Main {
  public static void main(String[] args) {
    Scanner scanner = new Scanner(System.in);

    int N = scanner.nextInt();
    long C = scanner.nextLong();
    long[] h = new long[N];
    for (int i = 0; i < N; i++) h[i] = scanner.nextInt();

    long[] dp = new long[N + 1];
    Deque<Line> deque = new LinkedList<>();

    for (int i = 1; i <= N; i++) {
      long hi = h[i - 1];
      while (deque.size() > 1) {
        Line last = deque.removeLast();
        if (deque.peekLast().y(hi) > last.y(hi)) {
          deque.addLast(last);
          break;
        }
      }
      if (!deque.isEmpty()) dp[i] = deque.peekLast().y(hi) + hi * hi + C;
      Line line = new Line(-2 * hi, hi * hi + dp[i]);
      while (deque.size() > 1) {
        Line first = deque.removeFirst();
        Line next = deque.peekFirst();
        if ((first.a - next.a) * (line.b - first.b) < (line.a - first.a) * (first.b - next.b)) {
          deque.addFirst(first);
          break;
        }
      }
      deque.addFirst(line);
    }

    System.out.println(dp[N]);
  }

  private static class Line {
    public final long a;
    public final long b;

    private Line(long a, long b) {
      this.a = a;
      this.b = b;
    }

    public long y(long x) {
      return x * a + b;
    }
  }
}