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: Time Limit Exceeded)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <string>
#include <cmath>
#include <map>
#include <unordered_map>
#include <set>
#include <queue>
#include <deque>

using namespace std;

int64_t C;

long f(int pos, const vector<int>& h, vector<int64_t>& res) {
  if (pos == 0) {
    return 0;
  }

  if (res[pos] == -1) {
    int64_t out = INT64_MAX;
    for (int i = pos - 1; i >= 0; i--) {
      int64_t diff = h[i] - h[pos];
      out = min(out, f(i, h, res) + diff * diff + C);
    }
    res[pos] = out;
  }

  return res[pos];
}

int main() {
  int n;
  cin >> n;
  cin >> C;

  vector<int> h(n);
  for (int i = 0; i < n; i++) {
    cin >> h[i];
  }

  vector<int64_t> res(n, -1);

  cout << f(n - 1, h, res) << endl;

  return 0;
}