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: Time Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;
#define ll long long

 ll n;
ll c;
 ll dp[100005];

ll solve(ll a[])
{
	dp[0] = 0;
  	dp[1] = dp[0]+abs(a[1]-a[0])*abs(a[1]-a[0]) +c;
  for(ll i=2;i<n;i++)
  {
    dp[i] = abs(a[i]-a[i-1])*abs(a[i]-a[i-1])+c+dp[i-1];
    for(ll j=1;j<n;j++)
    {
      if(i<j) break;
      if(dp[i]>(abs(a[i]-a[i-j])*abs(a[i]-a[i-j])+c+dp[i-j]))
      {
        dp[i] = abs(a[i]-a[i-j])*abs(a[i]-a[i-j])+c+dp[i-j];
        //cout<<a[i]<<"  "<<dp[i]<<endl;
      }
    } 
    
  } 
  return dp[n-1];
}
  

int main()
{
  cin>>n>>c;
 ll a[n];
  for(ll i=0;i<n;i++) cin>>a[i];
  
  cout<<solve(a)<<endl;
  return 0;
}