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>

#define endl '\n'
#define fi first
#define se second
#define fast ios_base::sync_with_stdio(0);cin.tie(0);
#define ALL(v) v.begin(), v.end()
#define ALLR(v) v.rbegin(), v.rend()
#define deb cout<<"HOLA\n";
#define UNIQUE(v) (v).resize(unique(ALL(v))-(v).begin())
#define forn(i,n) for (ll i = 0; i < n; i++)
#define forr(i,a,b) for (ll i = a; i <= b; i++)
#define bs(v,f) binary_search(ALL(v),f)
#define lb(v,f) lower_bound(ALL(v),f)
#define ub(v,f) upper_bound(ALL(v),f)
#define pb push_back

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef string str;
typedef long double ld;
typedef pair<int, int> ii;
typedef pair<ii, int> iii;
typedef vector<int> vi;
typedef vector<vector<int> > vvi;
typedef vector<ll> vl;
typedef vector<ii> vii;

vector<ll> a;ll n,k;
vector<__int128> dp(200005,-1);
__int128 ans(__int128 x){
    if(x==n-1)return 0;
    if(x>n-1)return 1e21;
    if(dp[x]!=-1)return dp[x];
    __int128 a1,mini=1e21;
    forn(i,n){
        a1=ans(x+i+1)+(a[x+i+1]-a[x])*(a[x+i+1]-a[x])+k;
        mini=min(a1,mini);
    }
    return dp[x]=mini;
}
void imp(__int128 x){
    stack<int>xd;
    while(x){
        xd.push(x%10);
        x/=10;
    }
    while(!xd.empty()){
        cout<<xd.top();
        xd.pop();
    }
}
void solve(){
	cin>>n>>k;
    a.resize(n+5);
    forn(i,n)cin>>a[i];
    imp(ans(0));
}
int main(){
    fast
	/*int t;
	cin>>t;
	while(t--)*/
		solve();
	return 0;
}