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 5.4.1) (judge verdict: Accepted)

#define f(I,H)(p[I]+h[I]*(h[I]-2*H))
long C,h[2<<17],p[2<<17],X,a;
P[2<<17];
s,t=-1,i,x,y;
main(){
	for(scanf("%*d%ld",&C);~scanf("%ld",&a);){
		for(;X=f(P[s],a),s<t&&X>=f(P[s+1],a);s++);
		h[i]=a;
		X=p[i]=i?a*a+C+X:0;
		for(;s<t&&(h[y=P[t]]-h[x=P[t-1]])*(f(y,0)-f(i,0))>=(a-h[y])*(f(x,0)-f(y,0));t--);
		P[++t]=i++;
	}
	printf("%ld",X);
}