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<iostream>
#include<vector>
using namespace std;
typedef long long ll;

struct line{
    ll a, b;
    bool operator==(const line other)const{
        return a==other.a && b==other.b;
    }
};

ll f(line l, ll x){
    return l.a*x + l.b;
}

const ll INF = 1ll<<60;

struct LiChaoTree{
    int n;
    vector<ll> xs;
    vector<line> v;
    line def{0, INF};
    LiChaoTree(vector<ll> x) : xs(x){
        n = 1;
        while(n < xs.size())  n *= 2;
        while(xs.size() < n)    xs.push_back(INF);
        v.resize(2*n+1, def);
    }
    void add_line(line p, int k=0, int l=0, int r=-1){
        if(r < 0)   r = n;
        int m = (l+r)/2;
        if(v[k] == def){
            v[k] = p;
            return;
        }
        ll lx = xs[l], mx = xs[m], rx = xs[r-1];
        bool left = f(p,lx) < f(v[k], lx);
        bool mid = f(p,mx) < f(v[k], mx);
        bool right = f(p,rx) < f(v[k], rx);
        if(left == right){
            if(left)    v[k] = p;
            return;
        }
        if(mid){
            swap(v[k], p);
        }
        // v[k]とpが交わっている部分について更新していくと考えるとわかりやすい
        if(left != mid){
            add_line(p, 2*k+1, l, m);
        }else{
            add_line(p, 2*k+2, m, r);
        }
    }
    ll query(int k){
        ll x = xs[k];
        k += n-1;
        ll ret = INF;
        while(true){
            if(!(v[k]==def)) ret = min(ret, f(v[k], x));
            if(k==0)    break;
            k = (k-1)/2;
        }
        return ret;
    }
};

int main(){
    int n;
    ll c;
    cin >> n >> c;
    vector<ll> h(n);
    for(int i = 0; i < n; i++)  cin >> h[i];
    LiChaoTree lct(h);
    vector<ll> dp(n, INF);
    dp[0] = 0;
    lct.add_line(line({-2*h[0], h[0]*h[0]+dp[0]}));
    for(int i = 1; i < n; i++){
        dp[i] = lct.query(i)+h[i]*h[i]+c;
        lct.add_line(line({-2*h[i], h[i]*h[i]+dp[i]}));
    }
    cout << dp[n-1] << endl;
    return 0;
}