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 PyPy3 (2.4.0) (judge verdict: Accepted)


class ConvexHullTrick:
	length = 0

	def __init__(self,size):
		self.A = [0]*size
		self.B = [0]*size

	def check(self,a,b):
		lhs = (self.B[self.length-2]-self.B[self.length-1])*(a-self.A[self.length-1])
		rhs = (self.B[self.length-1]-b)*(self.A[self.length-1]-self.A[self.length-2])
		return lhs>=rhs

	# 直線の傾きaは単調現象であること
	# Add a line y=a*x+b
	def add(self,a,b):
		while 2<=self.length and self.check(a,b):
			self.length -= 1
		self.A[self.length] = a
		self.B[self.length] = b
		self.length += 1
		return
	# min_i {a[i]*x+b[i]}
	def query(self,x):
		l,r = -1, self.length-1
		while l+1<r:
			m = (l+r)//2
			if self.eval(m+1,x)<=self.eval(m,x):
				l = m
			else:
				r = m
		return self.eval(r,x)

	def eval(self,k,x):
		return self.A[k]*x + self.B[k]

N,C = map(int,input().split())
h = list(map(int,input().split()))
#print(h)

cht = ConvexHullTrick(N)

dp = [0]*N
dp[0] = 0
cht.add(-2*h[0],h[0]**2+dp[0])
for i in range(1,N):
	#best = 99999999999999
	# for j in range(i):
	# 	#cost = dp[j]+(h[i]-h[j])**2+C
	# 	cost = (-2*h[j])*h[i]+(h[j]**2+dp[j])
	# 	best = min(best,cost)
	best = cht.query(h[i])
	dp[i] = best+h[i]**2+C

	cht.add(-2*h[i],h[i]**2+dp[i])

ans = dp[N-1]
print(ans)