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 pb push_back
#define loop(a, b, inc) for (int i = a; i <= b; i += inc)
typedef pair<int, int> pii;
typedef pair<int, string> ps;
typedef unsigned long long int llu;
typedef long long int ll;
typedef unsigned u;
const int MOD = 1e7+9, MAXN = 1e6;

vector<ll> h;
int n;
llu dp[MAXN], c;

llu solve(int i, int k) 
{
	if (i == n-1) return 0;
	if (dp[i] == -1) {
		dp[i] = (h[i]-h[i+1])*(h[i]-h[i+1]) + c + solve(i+1, k);
		for (int j = 2; j <= k && i+j <= n-1; ++j) {
			dp[i] = min(dp[i], (h[i]-h[i+j])*(h[i]-h[i+j])+c+solve(i+j, k));
		}
	}
	return dp[i];
} 

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	int aux;
	cin >> n >> c;
	loop(1, n, 1) {
		cin >> aux;
		h.pb(aux);
	}
	memset(dp, -1, sizeof(dp));
	cout << solve(0, n-1) << endl;
	
	return 0;
}