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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair
#define mod 1000000007
#define For(i,n) for(int i=0;i<n;i++)
#define Fre(i,a,b) for(int i = a; i < b; i++)
#define sf(n) scanf("%d", &n)
#define sff(a,b) scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pfn(n) printf("%d\n", n)
#define pfs(n) printf("%d ", n)
#define ff first
#define ss second
#define mem(a,b) memset(a,b,sizeof(a))
#define READ freopen("in.txt", "r", stdin)
#define WRITE freopen("out.txt", "w", stdout)
#define sz size()
#define number_of_digit(n) __builtin_ffs(n);
#define ll long long
#define T int
#define ld long double
inline void print(T n)
{
	if (n == 0)
	{
		putchar('0');
		putchar('\n');
	}
	else if (n == -1)
	{
		putchar('-');
		putchar('1');
		putchar('\n');
	}
	else
	{
		char buf[11];
		buf[10] = '\n';
		int i = 9;
		while (n)
		{
			buf[i--] = n % 10 + '0';
			n /= 10;
		}
		while (buf[i] != '\n')
			putchar(buf[++i]);
	}
}
int read()
{
       int cc = getc(stdin);
       for (;cc < '0' || cc > '9';)  cc = getc(stdin);
       int ret = 0;
       for (;cc >= '0' && cc <= '9';)
       {
          ret = ret * 10 + cc - '0';
          cc = getc(stdin);
       }
      return ret;
}
ll power(ll num,ll g)
{
	if(g==0)return 1;

	if(g%2==1)return (num*power((num*num)%mod,g/2))%mod;
	return power((num*num)%mod,g/2);
}
int main()
{
	ios_base::sync_with_stdio(false);
  	cin.tie(NULL);
  	cout.tie(NULL);

  	ll n,c;
  	cin>>n>>c;
  	ll a[n];
  	For(i,n)cin>>a[i];
  	ll b[n];
  	b[0]=0;
  	ll prev=0;
  	for(ll i=1;i<n;i++)
  	{
  		ll start=0,end=i-1;
  		while(start<end)
  		{
  			ll f=(start+end)/2;
  			ll g=f+1;
  			if((a[i]-a[f])*(a[i]-a[f])+b[f]<(a[i]-a[g])*(a[i]-a[g])+b[g])
  			{
  				end=f;
  			}
  			else start=g;
  		}
  		if((a[i]-a[prev])*(a[i]-a[prev])+b[prev]<(a[i]-a[start])*(a[i]-a[start])+b[start])start=prev;
  		else prev=start;
  		//cout<<i<<" "<<start<<" ";
  		b[i]=c+(a[start]-a[i])*(a[start]-a[i])+b[start];
  		//cout<<b[i]<<endl;
  	}
  	cout<<b[n-1]<<endl;
	return 0;
}