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 Rust (1.15.1) (judge verdict: Accepted)

use std::io::Read;

fn run() {
    let mut s = String::new();
    std::io::stdin().read_to_string(&mut s).unwrap();
    let mut it = s.trim().split_whitespace();
    let n: usize = it.next().unwrap().parse().unwrap();
    let c: i64 = it.next().unwrap().parse().unwrap();
    let mut h = vec![0i64; n];
    let mut dp = vec![0i64; n];
    h[0] = it.next().unwrap().parse().unwrap();
    let mut deq = std::collections::VecDeque::new();
    deq.push_back((-2 * h[0], 0 + h[0] * h[0]));
    for i in 1..n {
        h[i] = it.next().unwrap().parse().unwrap();
        let f = |(a, b)| a * h[i] + b;
        while deq.len() >= 2 {
            let a = *deq.get(0).unwrap();
            let b = *deq.get(1).unwrap();
            if f(a) < f(b) {
                break;
            }
            deq.pop_front();
        }
        dp[i] = f(*deq.front().unwrap()) + c + h[i] * h[i];
        let p = (-2 * h[i], dp[i] + h[i] * h[i]);
        while deq.len() >= 2 {
            let k = deq.len();
            let a = *deq.get(k - 2).unwrap();
            let b = *deq.get(k - 1).unwrap();
            if (b.1 - a.1) * (b.0 - p.0) < (p.1 - b.1) * (a.0 - b.0) {
                break;
            }
            deq.pop_back();
        }
        deq.push_back(p);
    }
    println!("{}", dp[n - 1]);
}

fn main() {
    run();
}