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 <cstdio>
#include <queue>

using namespace std;

long long h[200000];
long long dp[200000];

int main() {
    int n, i;
    long long c;
    deque <int> d;
    
    scanf("%d %lld", &n, &c);
    
    for (i = 0; i < n; i++) scanf("%lld", &h[i]);
    
    d.push_back(0);
    
    for (i = 1; i < n; i++) {
        while (d.size() >= 2) {
            int x = d[0];
            int y = d[1];
            
            if (dp[x] + (h[i] - h[x]) * (h[i] - h[x]) < dp[y] + (h[i] - h[y]) * (h[i] - h[y])) break;
            
            d.pop_front();
        }
        
        dp[i] = dp[d[0]] + (h[i] - h[d[0]]) * (h[i] - h[d[0]]) + c;
        
        d.push_back(i);
    }
    
    printf("%lld\n", dp[n - 1]);
    
    return 0;
}