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: Time Limit Exceeded)


def s0():return input()
def s1():return input().split()
def s2(n):return [input() for x in range(n)]
def s3(n):return [input().split() for _ in range(n)]
def s4(n):return [[x for x in s] for s in s2(n)]
def n0():return int(input())
def n1():return [int(x) for x in input().split()]
def n2(n):return [int(input()) for _ in range(n)]
def n3(n):return [[int(x) for x in input().split()] for _ in range(n)]
def t3(n):return [tuple(int(x) for x in input().split()) for _ in range(n)]
def p0(b,yes="Yes",no="No"): print(yes if b else no)
# from sys import setrecursionlimit
# setrecursionlimit(1000000)
# from collections import Counter,deque,defaultdict
# import itertools
# import math
# import networkx as nx
# from bisect import bisect_left,bisect_right
# from heapq import heapify,heappush,heappop
n,c=n1()
H=n1()

dp=[float("inf")]*n
dp[0]=0
for i in range(0,n):
    for j in range(1,n):
        if i+j<n:
            dp[i+j]=min(dp[i+j],dp[i]+(H[i]-H[i+j])**2+c)

print(dp[n-1])