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 (7.3.0) (judge verdict: Wrong Answer)

# -*- coding: utf-8 -*-

#############
# Libraries #
#############

import sys
input = sys.stdin.readline

import math
#from math import gcd
import bisect
import heapq
from collections import defaultdict
from collections import deque
from collections import Counter
from functools import lru_cache

#############
# Constants #
#############

MOD = 10**9+7
INF = float('inf')
AZ = "abcdefghijklmnopqrstuvwxyz"

#############
# Functions #
#############

######INPUT######
def I(): return int(input().strip())
def S(): return input().strip()
def IL(): return list(map(int,input().split()))
def SL(): return list(map(str,input().split()))
def ILs(n): return list(int(input()) for _ in range(n))
def SLs(n): return list(input().strip() for _ in range(n))
def ILL(n): return [list(map(int, input().split())) for _ in range(n)]
def SLL(n): return [list(map(str, input().split())) for _ in range(n)]


#####Shorten#####
def DD(arg): return defaultdict(arg)

#####Inverse#####
def inv(n): return pow(n, MOD-2, MOD)

######Combination######
kaijo_memo = []
def kaijo(n):
  if(len(kaijo_memo) > n): return kaijo_memo[n]
  if(len(kaijo_memo) == 0): kaijo_memo.append(1)
  while(len(kaijo_memo) <= n): kaijo_memo.append(kaijo_memo[-1] * len(kaijo_memo) % MOD)
  return kaijo_memo[n]

gyaku_kaijo_memo = []
def gyaku_kaijo(n):
  if(len(gyaku_kaijo_memo) > n): return gyaku_kaijo_memo[n]
  if(len(gyaku_kaijo_memo) == 0): gyaku_kaijo_memo.append(1)
  while(len(gyaku_kaijo_memo) <= n): gyaku_kaijo_memo.append(gyaku_kaijo_memo[-1] * pow(len(gyaku_kaijo_memo),MOD-2,MOD) % MOD)
  return gyaku_kaijo_memo[n]

def nCr(n,r):
  if n == r: return 1
  if n < r or r < 0: return 0
  ret = 1
  ret = ret * kaijo(n) % MOD
  ret = ret * gyaku_kaijo(r) % MOD
  ret = ret * gyaku_kaijo(n-r) % MOD
  return ret

######Factorization######
def factorization(n):
  arr = []
  temp = n
  for i in range(2, int(-(-n**0.5//1))+1):
    if temp%i==0:
      cnt=0
      while temp%i==0: 
        cnt+=1 
        temp //= i
      arr.append([i, cnt])
  if temp!=1: arr.append([temp, 1])
  if arr==[]: arr.append([n, 1])
  return arr

#####MakeDivisors######
def make_divisors(n):
  divisors = []
  for i in range(1, int(n**0.5)+1):
    if n % i == 0:
      divisors.append(i)
      if i != n // i: 
        divisors.append(n//i)
  return divisors

#####MakePrimes######
def make_primes(N):
  max = int(math.sqrt(N))
  seachList = [i for i in range(2,N+1)]
  primeNum = []
  while seachList[0] <= max:
    primeNum.append(seachList[0])
    tmp = seachList[0]
    seachList = [i for i in seachList if i % tmp != 0]
  primeNum.extend(seachList)
  return primeNum

#####GCD#####
def gcd(a, b):
    while b: a, b = b, a % b
    return a

#####LCM#####
def lcm(a, b):
    return a * b // gcd (a, b)

#####BitCount#####
def count_bit(n):
  count = 0
  while n:
    n &= n-1
    count += 1
  return count

#####ChangeBase#####
def base_10_to_n(X, n):
  if X//n: return base_10_to_n(X//n, n)+[X%n]
  return [X%n]

def base_n_to_10(X, n):
  return sum(int(str(X)[-i-1])*n**i for i in range(len(str(X))))

def base_10_to_n_without_0(X, n):
  X -= 1
  if X//n: return base_10_to_n_without_0(X//n, n)+[X%n]
  return [X%n]

#####IntLog#####
def int_log(n, a):
  count = 0
  while n>=a:
    n //= a
    count += 1
  return count

#############
# Main Code #
#############

N,C = IL()
A = IL()

dp = [0]

def cost(i,x):
  return dp[N-i-2-x] + (A[i+1+x]-A[i])**2 + C

def is_ok(i,mid1,mid2):
  if cost(i,mid1)>cost(i,mid2):
    return True
  else:
    return False

def meguru_bisect(i,ng, ok):
  if ok - ng == 2:
    return cost(i,ng+1)
  while (abs(ok - ng) > 2):
    mid1 = (ok*2 + ng) // 3
    mid2 = (ok + ng*2) // 3
    if is_ok(i,mid1,mid2):
      ok = mid1
    else:
      ng = mid2
  return min(cost(i,mid1),cost(i,mid2))

for i in range(N-1)[::-1]:
  dp.append(min((A[-1]-A[i])**2 + C,meguru_bisect(i,-1,N-1-i)))

print(dp[-1])