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 (GCC 5.4.1) (judge verdict: Accepted)

//Li Chao Segment Tree - trial
//dp(@)-Z
#include<stdio.h>
#define llinf 4154118101919364364

long long llmin(long long a,long long b){if(a<b){return a;}return b;}

//f(x) = (x-vertex)^2 + cv
typedef struct{
  long long vertex;
  long long cv;
}func;

long long fcalc(long long x,func f){
  return (x-f.vertex)*(x-f.vertex)+f.cv;
}

func li_chao[2097152];
long long ssize=1048576;

void sinit(){
  long long i;
  for(i=0;i<2097152;i++){
    li_chao[i].vertex=0;
    li_chao[i].cv=llinf;
  }
}

//add func f
//call:k=0,st=0,fi=0
void add(func f,long long k,long long st,long long fi){
  long long nst=fcalc(st,f),nfi=fcalc(fi-1,f);
  long long pst=fcalc(st,li_chao[k]),pfi=fcalc(fi-1,li_chao[k]);
  if(nst>=pst && nfi>=pfi){return;}
  if(nst<=pst && nfi<=pfi){
    li_chao[k]=f;
    return;
  }
  long long mid=(st+fi)/2;
  add(f,2*k+1,st,mid);  
  add(f,2*k+2,mid,fi);  
}

//get min f(x)
long long get(long long x){
  long long p=x+ssize-1,res=llinf;
  while(p>=0){
    res=llmin(fcalc(x,li_chao[p]),res);
    if(p==0){break;}
    p=(p-1)/2;
  }
  return res;
}

int main(){
  sinit();
  long long n,c,h[262144],i;
  func f;
  scanf("%lld%lld",&n,&c);
  for(i=0;i<n;i++){
    scanf("%lld",&h[i]);
    f.vertex=h[i];
    if(i==0){
      f.cv=c;
    }
    else{
      f.cv=get(h[i])+c;
    }
    add(f,0,0,ssize);
  }
  printf("%lld\n",get(h[n-1]));
}