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>
using namespace std;
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define line pair<long long,long long>

int n;
long long C, h[200007];
long long dp[200007];
const long long inf = 1e18;

class ConvexHullTrick{
public:
	deque <line> cur;
	bool bad(line a,line b,line c){
		return ((a.se-b.se)*(c.fi-a.fi)>(b.fi-a.fi)*(a.se-c.se));
	}
	void add(long long A,long long B){
		line p=mp(A,B);
		while((int)cur.size()>=2 && bad(cur[(int)cur.size()-2],cur[(int)cur.size()-1],p)){
			cur.pop_back();
		}
		cur.push_back(p);
	}
	long long query(long long x){
		if(cur.empty()) return inf;
		while((int)cur.size()>=2 && cur[0].fi*x+cur[0].se > cur[1].fi*x+cur[1].se) cur.pop_front();
		return cur[0].fi*x+cur[0].se;
	}
} slope;

int main(){
	ios_base::sync_with_stdio(0);
	cin >> n >> C;
	for(int i = 1; i <= n; i++){
		cin >> h[i];
	}
	dp[1] = 0;
	slope.add(-2LL * h[1], dp[1] + C + h[1] * h[1]);
	for(int i = 2; i <= n; i++){
		dp[i] = slope.query(h[i]) + h[i] * h[i];
		slope.add(-2LL * h[i], dp[i] + C + h[i] * h[i]);
	}
	cout << dp[n];
}