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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

int n;
long long c, dp[1000005], ar[1000005];
long long calc (int i, int j) {
	return dp[i] + (ar[j] - ar[i]) * (ar[j] - ar[i]) + c;
}
int main () {
	ios_base::sync_with_stdio (0);
  	cin.tie (0);
  	cin >> n >> c;
  	for (int i = 1; i <= n; i++) {
    	cin >> ar[i];
    }
  	dp[1] = 0;
  	dp[2] = calc (1, 2);
  	for (int i = 3; i <= n; i++) {
      	dp[i] = 1e18;
    	int L = 1, R = i - 1;
   		for (int qq = 1; qq <= 200; qq++) {
   		    int dis = (R - L) / 3;
        	int M1 = L + dis, M2 = R - dis;
          	if (calc (M1, i) <  calc (M2, i))
              R = M2;
          	else
              L = M1;
        }
      	for (int j = L; j <= R; j++) {
        	dp[i] = min (dp[i], calc (j, i)); 
        }
    }
  	cout << dp[n] << '\n';
}