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 <bits/stdc++.h>
#define rep(i,n)for(int i=0;i<(n);i++)
using namespace std;
typedef long long ll;
typedef pair<int,int>P;

struct line{
	ll a,b;
	ll get(ll x){
		return a*x+b;
	}
};
struct CHT{
	deque<line>que;
	bool check(line a,line b,line c){
		double A[]{a.a,b.a,c.a},B[]{a.b,b.b,c.b};
		return (B[1]-B[2])*(A[1]-A[0])<=(B[0]-B[1])*(A[2]-A[1]);
	}
	void add(line l){
		while(que.size()>1&&check(que[que.size()-2],que.back(),l))que.pop_back();
		que.push_back(l);
	}
	double get(double x){
		if(que.empty())return INFINITY;
		while(que.size()>1&&que[0].get(x)>=que[1].get(x))que.pop_front();
		return que[0].get(x);
	}
};

ll h[300000];
ll a[300000],b[300000];
ll dp[300000];

int main(){
	int n;ll C;cin>>n>>C;
	rep(i,n)scanf("%lld",&h[i]);
	CHT cht;
	dp[0]=0;
	cht.add({-2*h[0],h[0]*h[0]+C});
	for(int i=1;i<n;i++){
		dp[i]=cht.get(h[i])+h[i]*h[i];
		cht.add({-2*h[i],dp[i]+h[i]*h[i]+C});
	}
	cout<<dp[n-1]<<endl;
}