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;
const int N = 2e5 + 100; 
int n;
long long h[N], c;

struct Sol {
	long long m = 0;
	long long k = 0;

	Sol() = default;
	Sol(long long _m, long long _k) {
		m = _m;
		k = _k;
	}

	long long eval(long long x) {
		return m * x + k;
	}

	double to(Sol &s) {
		if (m == s.m) {
			return LLONG_MAX;
		}
		return (double) (s.k - k) / (m - s.m);
	}
};

Sol dq[N];

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

	cin >> n >> c;
	for (int i = 1; i <= n; i++) {
		cin >> h[i];
	}
	
	int l = 0, r = 0;
	long long opt = 0;
	for (int i = 2; i <= n; i++) {
		Sol nsol(-2 * h[i - 1], opt + c + h[i - 1] * h[i - 1]);

		while (r - l >= 2 && dq[r - 2].to(dq[r - 1]) >= dq[r - 2].to(nsol)) {
			r--;
		}
		dq[r++] = nsol;

		while (l + 1 < r && dq[l].eval(h[i]) >= dq[l + 1].eval(h[i])) {
			l++;
		}

		opt = dq[l].eval(h[i]) + h[i] * h[i];
	}

	cout << opt << '\n';

	return 0;
}