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 <bits/stdc++.h>
using namespace std;
# define MOD 1000000007
typedef long long ll;
class line{
public:
    ll m, c;
    line(ll m, ll c){
        this->m = m;
        this->c = c;
    }
    ll value(ll x){
        return x*m + c;
    }
};
ll intersection(line& l1, line& l2){
    return ceil(1.0*(l2.c - l1.c)/(l1.m - l2.m));
}
class CH{
    deque<pair<line, ll>> dq;
public:
    void insert(line l){
        if(dq.empty()){
            dq.push_back({l, 0});
        }else{
            while(!dq.empty()){
                ll x = intersection(l, dq.back().first);
                if(x <= dq.back().second){
                    dq.pop_back();
                }else break;
            }
            ll x = intersection(l, dq.back().first);
            dq.push_back({l, x});
        }
    }
    ll min_intercept(ll x){
        while(dq.size() > 1){
            if(dq[1].second <= x){
                dq.pop_front();
            }else break;
        }
        return dq[0].first.value(x);
    }
};
int main(){
    ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
    int n;
    ll C;
    cin>>n>>C;
    ll heights[n];
    for(int i=0;i<n;i++){
        cin>>heights[i];
    }
    ll dp[n];
    dp[0] = 0;
    CH lines;
    lines.insert(line(-2*heights[0], heights[0]*heights[0]));
    for(int i=1;i<n;i++){
        dp[i] = heights[i]*heights[i] + C + lines.min_intercept(heights[i]);
        lines.insert(line(-2*heights[i], dp[i] + heights[i]*heights[i]));
    }
    cout<<dp[n-1];
	return 0;
}