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<stdio.h>
#include<iostream>
#include<algorithm>
#include<math.h>
#include<queue>
#include<map>
#include<bitset>
#include<stack>
#include<set>
#include<vector>
#include <time.h>
#include<string.h>
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
typedef pair <ll, int> pli;
typedef pair <db, db> pdd;

const int maxn = 4e5+5;
const int Mod=1000000007;
const int INF = 0x3f3f3f3f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const double e=exp(1);
const db PI = acos(-1);
const db ERR = 1e-10;

#define Se second
#define Fi first
#define pb push_back
#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
pll que[maxn];
ll h[maxn];
ll dp[maxn];
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int n;ll c;
    scanf("%d%lld",&n,&c);
    for(int i=1;i<=n;i++) scanf("%lld",&h[i]);
    dp[1]=0;
    int tail=-1,head=0;
    que[++tail]=pll(h[1],1LL*h[1]*h[1]);
    for(int i=2;i<=n;i++)
    {
        while(head<tail&&(que[head+1].Se-que[head].Se)<=2LL*h[i]*(que[head+1].Fi-que[head].Fi)) head++;
        dp[i]=que[head].Se-que[head].Fi*que[head].Fi+1LL*(h[i]-que[head].Fi)*(h[i]-que[head].Fi)+c;
        while(head<tail&&1.0*(que[tail].Se-que[tail-1].Se)*(h[i]-que[tail].Fi)<1.0*(dp[i]-que[tail].Se)*(que[tail].Fi-que[tail-1].Fi)) tail--;
        que[++tail]=pll(h[i],dp[i]+h[i]*h[i]);
       // dbg2(i,dp[i]);
       // dbg2(head,tail);
    }
    printf("%lld\n",dp[n]);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}