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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

int n;
long long C;
int h[200055];

// hi^2 - 2hihj + hj^2 + c
// -2hihj + hj^2 + hi^2 + c

int deq[200055];
long long dp[200055];

inline long long f(int i, int j) {
	return 1ll * (h[i] - h[j]) * (h[i] - h[j]) + dp[j] + C;
}
inline bool check(int i, int j, int k) {
	long long a1 = -2ll * h[i], b1 = dp[i] + 1ll * h[i] * h[i];
	long long a2 = -2ll * h[j], b2 = dp[j] + 1ll * h[j] * h[j];
	long long a3 = -2ll * h[k], b3 = dp[k] + 1ll * h[k] * h[k];
	
	return 1. * (a2 - a1) * (b3 - b2) >= 1. * (b2 - b1) * (a3 - a2);
}
int main() {
	cin >> n >> C;
	
	for (int i = 1; i <= n; i++) {
		scanf("%d", &h[i]);
	}
	
	int s = 0, t = 1;
	deq[s] = 1;
	
	for (int i = 2; i <= n; i++) {
		while (s + 1 < t && f(i, deq[s]) >= f(i, deq[s + 1])) {
			s++;
		}
		
		dp[i] = f(i, deq[s]);
		
		while (s + 1 < t && check(deq[t - 2], deq[t - 1], i)) {
			t--;
		}
		
		deq[t++] = i;
	}
	
	printf("%lld\n", dp[n]);
}