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<iostream>
#include<cstdio>
#include<algorithm>
#include<cassert>
#include<cmath>
#include<vector>
#include<map>
#include<set>
#include<string>
#include<queue>
#include<stack>
using namespace std;
#define MOD 1000000007
#define MOD2 998244353
#define INF ((1<<30)-1)
#define LINF (1LL<<60)
#define EPS (1e-10)
typedef long long Int;
typedef pair<Int, Int> P;

Int n, c;
Int dp[220000];
Int h[220000];

Int calc(P f, Int x){
    return f.first * x + f.second;
}

struct CHTrick{
    vector<Int> to;
    vector<P> f;

    CHTrick(){}

    void add_line(Int a, Int b){
        P new_f = P(a, b);
        if(f.empty()){
            f.push_back(new_f);
            return;
        }
        if(calc(new_f, INF) >= calc(f.back(), INF))return;
        if(calc(new_f, -INF) <= calc(f.front(), -INF)){
            f.clear();
            to.clear();
            f.push_back(new_f);
            return;
        }

        while(!to.empty()){
            int lastl = to.back() + 1;
            if(calc(new_f, lastl) <= calc(f.back(), lastl)){
                f.pop_back();
                to.pop_back();
            }
            else break;                
        }

        int bottom;
        if(to.empty())bottom = -INF;
        else bottom = to.back() + 1; //f.back();
        int top = INF;//new_f

        while(top - bottom > 1){
            int mid = (top + bottom) / 2;
            if(calc(new_f, mid) <= calc(f.back(), mid))top = mid;
            else bottom = mid;
        }
        to.push_back(bottom);
        f.push_back(new_f);
    }
    
    Int get_min(int x){
        Int ind = lower_bound(to.begin(), to.end(), x) - to.begin();
        return calc(f[ind], x);
    }
};

int main(){
    cin >> n >> c;
    for(int i = 0;i < n;i++)cin >> h[i];
    CHTrick ch;
    
    dp[0] = 0;
    ch.add_line(-2*h[0], dp[0] + h[0] * h[0]);
    for(int i = 1;i < n;i++){
        dp[i] = ch.get_min(h[i]) + h[i] * h[i] + c;
        ch.add_line(-2*h[i], dp[i] + h[i] * h[i]);
    }
    cout << dp[n-1] << endl;    
    return 0;
}