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 (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int main()
{
	int i, N, h[200001];
	long long C;
	scanf("%d %lld", &N, &C);
	for (i = 1; i <= N; i++) scanf("%d", &(h[i]));
	
	int j, k, l[2] = {1, 2}, r[2] = {1, 2}, L, R, M;
	long long dp[200001] = {0, 0, (long long)(h[2] - h[1]) * (h[2] - h[1]) + C};
	for (i = 3; i <= N; i++) {
		L = l[0];
		R = r[0];
		while (L < R) {
			M = (L + R) / 2;
			if (dp[M] + (long long)(h[i] - h[M]) * (h[i] - h[M]) > dp[M+1] + (long long)(h[i] - h[M+1]) * (h[i] - h[M+1])) L = M + 1;
			else R = M;
		}
		dp[i] = dp[L] + (long long)(h[i] - h[L]) * (h[i] - h[L]) + C;
		
		L = l[1];
		R = r[1];
		while (L < R) {
			M = (L + R) / 2;
			if (dp[M] + (long long)(h[i] - h[M]) * (h[i] - h[M]) > dp[M+1] + (long long)(h[i] - h[M+1]) * (h[i] - h[M+1])) L = M + 1;
			else R = M;
		}
		if (dp[L] + (long long)(h[i] - h[L]) * (h[i] - h[L]) + C <= dp[i]) {
			dp[i] = dp[L] + (long long)(h[i] - h[L]) * (h[i] - h[L]) + C;
			l[0] = l[1];
			r[0] = r[1];
			l[1] = i;
			r[1] = i;
		} else r[1]++;
	}
	
	printf("%lld\n", dp[N]);
	fflush(stdout);
	return 0;
}