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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
using ll = unsigned long long;

ll sq(ll x){return x*x;}
ll n, c, h[200005], dp[200005];

int main(){

	cin.tie( nullptr );
	ios_base::sync_with_stdio(0);
	
	cin >> n >> c; ll rt = sqrt(c);
	for(int i=0; i<n; i++) cin >> h[i];
	for(int i=1; i<n; i++) dp[i] = LLONG_MAX;
	for(int i=0; i<n; i++){
		dp[i+1] = min(dp[i+1], c + dp[i] + sq(h[i+1]-h[i]));
		int j = upper_bound(h, h+n, h[i]+rt) - h - 1;
		int step=1000; while(j<n && (step--))
			dp[j] = min(dp[j], c + dp[i] + sq(h[j] - h[i])), j++;
	}cout << dp[n-1] << "\n";

	return 0;
}