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: Runtime Error)

#include <cstdio>
#include <exception>
#include <cstring>
#include <cassert>
#include <algorithm>
#include <string>
#include <iostream>
#include <vector>
#include <queue>
#include <map>

using namespace std;

const int CAP = 2e5;
const int N = 1e6;
const int T = 4e6;
const long long INF = 1LL << 60;

struct line {
	long long a;
	long long b;
	long long eval(long long x) {
		return a * x + b;
	}
};

bool vis[T];
line c[T];

void modify(int i, int l, int r, line a) {
	if (!vis[i]) {
		vis[i] = 1;
		c[i] = a;
		return;
	}
	int mid = (l + r) >> 1;
	if (a.eval(l) <= c[i].eval(l) && a.eval(r) <= c[i].eval(r)) {
		swap(c[i], a);
		return;
	}
	if (a.eval(l) >= c[i].eval(l) && a.eval(r) >= c[i].eval(r)) {
		return;
	}
	if (a.eval(l) < c[i].eval(l)) {
		swap(a, c[i]);
	}

	if (a.eval(mid) > c[i].eval(mid)) {
		modify(i * 2 + 1, mid + 1, r, a);
	} else {
		modify(i * 2, l, mid, a);
	}
}

long long value(int i, int l, int r, int x) {
	if (!vis[i]) {
		return INF;
	}
	if (l == r) {
		return c[i].eval(x);
	}
	long long ans = c[i].eval(x);
	int mid = (l + r) >> 1;
	if (x < mid) {
		return min(ans, value(2 * i, l, mid - 1, x));
	} else {
		return min(ans, value(2 * i + 1, mid, r, x));
	}
}

long long dp[N];

int main() {
	::ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int n;
	long long c;
	cin >> n >> c;
	for (int i = 0; i < n; ++i) {
		static long long h;
		cin >> h;
		if (i != 0) {
			dp[i] = value(1, 1, N, h) + c + h * h;
		}
		line cur = { -2 * h, dp[i] + h * h };
		modify(1, 1, N, cur);
		//cout << dp[i] << " ";
	}
	cout << dp[n - 1] << "\n";
	return 0;
}