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: Wrong Answer)

#include <bits/stdc++.h>

#define ff first
#define ss second
#define mp make_pair

using namespace std;

typedef long long ll;

bool D(pair<ll,ll> A, pair<ll,ll> B, pair<ll,ll> C) {
	return A.ff*B.ss + A.ss*C.ff + B.ff*C.ss -
	(B.ss*C.ff + A.ff*C.ss + A.ss*B.ff) > 0;
}

vector<pair<ll,ll>> v;

void add(pair<ll,ll> a) {
	while(v.size() >= 2 and D(v[v.size()-2], v[v.size()-1], a))
		v.pop_back();
	v.push_back(a);
}

ll dot(pair<ll,ll> a, pair<ll,ll> b) {
	return a.ff*b.ff + a.ss*b.ss;
}

double f(int m, int x) {
	pair<double, double> a = mp(v[m+1].ff-v[m].ff, v[m+1].ss-v[m].ss);
	pair<double, double> vec = mp(-a.ss, a.ff);
	double len = sqrt(vec.ff*vec.ff + vec.ss*vec.ss);
	vec.ff/=len; vec.ss/=len;
	return x*vec.ff + vec.ss;
}

ll bisearch(int x) {
	if(v.size() == 0) return 0;
	int l = 0, r = (int)v.size()-2;
	while(l < r) {
		int m = (l+r)/2;
		if(f(m, x) < f(m+1, x))
			r = m;
		else
			l = m+1;
	}

	if(v.size() == 1) return dot(mp(x,1), v[0]);
	return min(dot(mp(x,1), v[r]), dot(mp(x,1), v[r+1]));
}

int n, h[200005];
ll c, dp[200005];
bool mark[200005];

ll f() {
	for(int i = 0; i < n; i++) {
		if(i == 0) dp[i] = 0;
		else dp[i] = bisearch(h[i]) + (ll)h[i]*h[i] + c;
		add(mp(-2LL*h[i], (ll)h[i]*h[i] + dp[i]));
	}

	return dp[n-1];
}

int main() {
	scanf("%d%lld", &n, &c);

	for(int i = 0; i < n; i++)
		scanf("%d", &h[i]);

	printf("%lld\n", f());
	
    return 0;
}