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: Runtime Error)

#include<bits/stdc++.h>
using namespace std;

#define y1 hasjdnasd
#define ii pair < int , int >
#define iii pair < int , ii >
#define iv pair < ii , ii >

#define fi first
#define se second
#define pb push_back
#define fr front()

#define FOR(i , x , n) for(int i = x ; i <= n ; ++i)
#define FORD(i , x , n) for(int i = x ; i >= n ; --i)
#define REP(i , x) for(int i = 0 ; i < x ; ++i)

#define ll long long
#define oo 1e18 + 7
#define int long long

const int N = 1e5 + 5;
int n , C;
int h[N] , dp[N];

struct line
{
    int a , b;
    void operator -= (line const &o)
    {
        a -= o.a;
        b -= o.b;
    }
};

deque < line > dq;

int CCW(line l1 , line l2 , line l3)
{
    l3 -= l2;
    l2 -= l1;
    return l2.a * l3.b - l2.b * l3.a;
}

int val(line l , int x)
{
    return l.a * x + l.b;
}

void addline(line l)
{
    while(dq.size() >= 2 && CCW(dq[dq.size() - 2] , dq[dq.size() - 1] , l) >= 0)
        dq.pop_back();
    dq.pb(l);
}

int get(int x)
{
    int l = 0 , r = dq.size() - 1 , mid = 0;
    while(l <= r)
    {
        mid = (l + r) / 2;
        if(mid != 0 && val(dq[mid - 1] , x) < val(dq[mid] , x)) r = mid - 1;
        else if(mid != dq.size() - 1 && val(dq[mid + 1] , x) < val(dq[mid] , x)) l = mid + 1;
        else break;
    }
    return val(dq[mid] , x);
}

main()
{
    //freopen("Z.inp","r",stdin);
    ios::sync_with_stdio(0);
    cin.tie(0);
    cin >> n >> C;
    FOR(i , 1 , n)
        cin >> h[i];
    line l;
    l.a = -2 * h[1];
    l.b = h[1] * h[1];
    addline(l);
    FOR(i , 2 , n)
    {
        dp[i] = h[i] * h[i] + get(h[i]) + C;
        l.a = -2 * h[i];
        l.b = dp[i] + h[i] * h[i];
        addline(l);
    }
    cout << dp[n];
}