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: Accepted)

#include "bits/stdc++.h"
using namespace std;
using ll = long long;

struct line
{
	ll a, b;
	  line(ll m, ll c)
	{
		a = m;
		b = c;
	}
	ll operator[] (int x)
	{
		return a * x + b;
	}
	double operator*(line const &o)
	{
		return double (b - o.b) / (o.a - a);
	}
};

int main()
{
	cin.tie(0);
	cin.sync_with_stdio(0);

	int n;
	ll c;
	cin >> n >> c;
	deque < pair < line, ll > >dp;
	for (int i = 0; i < n; ++i)
	{
		ll x;
		cin >> x;
		while (dp.size() > 1 && dp[0].first[x] >= dp[1].first[x])
			dp.pop_front();
		auto v = i ? dp.front().first[x] + (x * x + c) : 0;
		auto l = line(-2 * x, v + x * x);
		while (dp.size() > 1
			   && (dp.back().first * dp[dp.size() - 2u].first) >= (l * dp[dp.size() - 2u].first))
			dp.pop_back();
		dp.emplace_back(l, v);
	}
	cout << dp.back().second;
}