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 Lichaotree:
    #区間は1-indexed
    #min
    def __init__(self, X):
        self.N = len(X)
        self.N0 = 2**(self.N-1).bit_length()
        self.comp = {v : k for k, v in enumerate(X)}
        self.inf = 10**12
        self.X = X + [self.inf]*(self.N0 - self.N)
        self.data = [None]*(2*self.N0)
        self.size = [0]*self.N0 + [1]*self.N0
        for i in range(self.N0-1, 0, -1):
            self.size[i] = 2*self.size[2*i]
        self.start = [0]*(2*self.N0)
        for i in range(1, 2*self.N0):
            if i == i&(-i):
                continue
            self.start[i] = self.start[i-1] + self.size[i]
        
    def f(self, line, x):
        return line[0]*x + line[1]

    def _update(self, k, line):
        
        if self.data[k] is None:
            self.data[k] = line
            return
        
        l = self.start[k]
        r = l + self.size[k]        
        
        while True:
            if self.data[k] is None:
                self.data[k] = line
                return
            
            m = (l+r)//2
            lx = self.X[l]
            mx = self.X[m]
            rx = self.X[r-1]
            gl = (self.f(self.data[k], lx) > self.f(line, lx))
            gm = (self.f(self.data[k], mx) > self.f(line, mx))
            gr = (self.f(self.data[k], rx) > self.f(line, rx))
            if gl and gr:
                self.data[k] = line
                return
            
            if not gl and not gr:
                return
            
            if gm:
                self.data[k], line = line, self.data[k]
                gl = not gl
                
            if gl:
                r = m
                k = 2*k
            else:
                l = m
                k = 2*k+1
    
    def query(self, x):
        k = self.comp[x] + self.N0
        s = self.inf
        while k > 0:
            if self.data[k]:
                s = min(s, self.f(self.data[k], x))
            k = k >> 1
        return s
            
    def addline(self, line):
        self._update(1, line)
    
    def addlineseg(self, l, r, line):
        #区間[l, r)にlineを追加,l, rはXの元
        
        L = l+self.N0
        R = r+self.N0
        while L < R:
            if R & 1:
                R -= 1
                self._update(R, line)
            if L & 1:
                self._update(L, line)
                L += 1
            L >>= 1
            R >>= 1

N, C = map(int, input().split())
H = list(map(int, input().split()))

T = Lichaotree(H)
dp = [10**19]*N
dp[0] = 0
T.addline((-2*H[0], H[0]**2))
for i in range(1,N):
    mi = T.query(H[i])
    dp[i] = mi + C + H[i]**2
    T.addline((-2*H[i], H[i]**2+dp[i]))
print(dp[-1])