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: Accepted)

#include <iostream>
#include <cmath>
#include <cstring>
#include <algorithm>
#define  ll long long
#define size 200020
using namespace std;
ll a[size],q[size];
ll dp[size];
double slope(int j,int k) {
	double y=dp[j]+a[j]*a[j]-dp[k]-a[k]*a[k];
	double x=a[j]-a[k];
	return y/x;
}
int main()
{
	int n;
	ll c;
	while(cin>>n>>c){
		for(int i=1;i<=n;i++)
			cin>>a[i];
		dp[0]=0;
		int head=1,tail=1;
		q[head]=1;
		for(int i=2;i<=n;i++){
			while(head<tail&&slope(q[head],q[head+1])<=2*a[i]) head++;
			int j=q[head];
			dp[i]=dp[j]+c+(a[i]-a[j])*(a[i]-a[j]);
			while(head<tail&&slope(q[tail-1],q[tail])>=slope(q[tail],i)) tail--;
			q[++tail]=i;
//			cout<<dp[i]<<endl;
		}
		cout<<dp[n]<<endl;
	}
}