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)

#include<stdio.h>
#include<stdlib.h>

typedef long long int int64;

typedef struct line{
  int64 a,b;
} line;

int64 func(line l,int x){
  return l.a*x+l.b;;
}

void run(void){
  int n;
  int64 c;
  scanf("%d%lld",&n,&c);
  int *h=(int *)calloc(n,sizeof(int));
  int i;
  for(i=0;i<n;i++) scanf("%d",h+i);
  int64 *dp=(int64 *)calloc(n,sizeof(int64));
  line *q=(line *)calloc(n,sizeof(line));
  int front=0,last=0;
  q[last++]=(line){-2*h[0],(int64)h[0]*h[0]};
  for(i=1;i<n;i++){
    const int x=h[i];
    while(front+1<last && func(q[front],x)>=func(q[front+1],x)) front++;
    dp[i]=(int64)x*x+c+func(q[front],x);
    const line l={-2*x,dp[i]+(int64)x*x};
    while(front+1<last){
      const line a=q[last-2];
      const line b=q[last-1];
      if((l.a-b.a)*(a.b-l.b)<(l.a-a.a)*(b.b-l.b)) break;
      last--;
    }
    q[last++]=l;
  }
  printf("%lld\n",dp[n-1]);
}

int main(void){
  run();
  return 0;
}