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: Wrong Answer)

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

    long[] dp = new long[N];
    for (int i = 1; i < N; i++) {
      int left = 0;
      int right = i;
      while (right - left > 1) {
        int mid = (left + right) / 2;
        long m1 = dp[mid - 1] + cost(i, mid - 1, h, C);
        long m2 = dp[mid] + cost(i, mid, h, C);
        if (m1 > m2) {
          left = mid;
        } else {
          right = mid;
        }
      }
      dp[i] = dp[left] + cost(i, left, h, C);
    }
    System.out.println(dp[N - 1]);
  }

  private static long cost(int i, int j, int[] h, long C) {
    return (long) (h[i] - h[j]) * (h[i] - h[j]) + C;
  }
}