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 <bits/stdc++.h>
#define LL long long
#define PII pair<int,int>
using namespace std;
const int MAXN = 100005;
const LL MOD = 998244353;

LL N, C, s, t, h[MAXN], dp[MAXN];
int deq[MAXN];

bool useless(int a, int b, int c) {
	LL m1 = -2 * h[a], b1 = dp[a] + h[a] * h[a];
	LL m2 = -2 * h[b], b2 = dp[b] + h[b] * h[b];
	LL m3 = -2 * h[c], b3 = dp[c] + h[c] * h[c];
	if ((b2 - b3) / (m3 - m2) < (b1 - b2) / (m2 - m1)) return true;
	if ((b2 - b3) / (m3 - m2) > (b1 - b2) / (m2 - m1)) return false;
	return (b2 - b3) % (m3 - m2) * (m2 - m1) <= (b1 - b2) % (m2 - m1) * (m3 - m2);
}

LL f(int j, LL x) {
	return dp[j] + h[j] * h[j] - 2 * h[j] * x;
}

int main() {
	ios_base::sync_with_stdio(0); cin.tie(0);
	
	cin >> N >> C;
	for (int i = 1; i <= N; i++) cin >> h[i];
	
	dp[1] = 0;
	for (int i = 2; i <= N; i++) {
		while (t - s >= 2 && useless(deq[t - 2], deq[t - 1], i - 1)) t--;
		deq[t++] = i - 1;
		
		while (t - s >= 2 && f(deq[s], h[i]) >= f(deq[s + 1], h[i])) s++;
		dp[i] = f(deq[s], h[i]) + h[i] * h[i] + C;
	}
	
	cout << dp[N] << "\n";
}