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 <cstdio>
#include <cstring>
#include <iostream>
#include <string>
#include <cmath>
#include <bitset>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <algorithm>
#include <complex>
#include <unordered_map>
#include <unordered_set>
#include <random>
#include <cassert>
using namespace std;
typedef long long int ll;
typedef pair<ll, ll> P;
int main()
{
	int n; ll c; cin>>n>>c;
	ll h[200001];
	for(int i=1; i<=n; i++) cin>>h[i];
	deque<P> line;
	ll dp=0;
	for(int i=1; i<=n; i++){
		if(i>1){
			while(line.size()>1){
				ll a1=line[0].first, b1=line[0].second;
				ll a2=line[1].first, b2=line[1].second;
				if(a1*h[i]+b1<a2*h[i]+b2) break;
				line.pop_front();
			}
			ll a1=line[0].first, b1=line[0].second;
			dp=a1*h[i]+b1+h[i]*h[i]+c;
		}
		if(i==n) break;
		ll a0=-2*h[i], b0=h[i]*h[i]+dp;
		if(!(line.size()>=1 && line.back().first==a0 && line.back().second<=b0)){
			while(line.size()>1){
				ll a1=line[line.size()-1].first, b1=line[line.size()-1].second;
				ll a2=line[line.size()-2].first, b2=line[line.size()-2].second;
				if((b1-b2)*(a1-a0)<(b0-b1)*(a2-a1)) break;
				line.pop_back();
			}
			line.push_back(P(a0, b0));
		}
	}
	cout<<dp<<endl;
	return 0;
}