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>
#include<stdint.h>
#include<inttypes.h>

typedef int32_t i32;
typedef int64_t i64;

typedef struct line {
  i64 a, b;
} line;

i64 eval (line l, i64 x) {
  return l.a * x + l.b;
}

int cond (line p, line q, line r) {
  return -(p.b - q.b) * (q.a - r.a) < -(q.b - r.b) * (p.a - q.a);
}

#define ALLOC(size,type) ((type*) calloc ((size), sizeof (type)))

void run (void) {
  i32 n;
  i64 c;
  scanf ("%" SCNi32 "%" SCNi64, &n, &c);
  i32 *h = ALLOC (n, i32);
  for (i32 i = 0; i < n; ++i) {
    scanf ("%" SCNi32, h + i);
  }
  i64 *dp = ALLOC (n, i64);
  line *d = ALLOC (n, line);
  i32 front = 0;
  i32 last = 0;
  d[last++] = (line) {-2 * h[0], (i64)h[0] * h[0] + c};
  for (i32 i = 1; i < n; ++i) {
    i32 x = h[i];
    for (; front < last - 1 && eval (d[front], x) >= eval (d[front + 1], x); ++front);
    dp[i] = eval (d[front], x) + (i64) x * x;
    line l = {-2 * h[i], dp[i] + (i64) x * x + c};
    for (; front + 1 < last && !cond (d[last - 2], d[last - 1], l); --last);
    d[last++] = l;
  }
  printf ("%" PRIi64 "\n", dp[n - 1]);
}

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