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>
#include<iostream>
#include<cmath>
#include<vector>
#include<string>
#include<queue>
#include<set>
#include<map>
#include<algorithm>
#include<functional>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<ll, ll> pll;
typedef pair<ull, ull> pullull;
typedef pair<ll, int> plli;
typedef pair<int, pii> pipii;
typedef vector<vector<int> > mati;
typedef vector<vector<double> > matd;
typedef vector<ll> vll;
typedef vector<vector<ll>> vvll;
typedef vector<vector<vector<ll>>> vvvll;
typedef vector<bool> vb;
typedef vector<vector<bool>> vvb;
typedef vector<vector<vector<bool>>> vvvb;

#define FOR(i,x,y) for(ll i=(ll)x; i<(ll)y; ++i)
#define REP(i,y) FOR(i, 0, y)

template <typename T>
void vec_print(vector<T> VEC){
  REP(i, VEC.size()){
    cout << VEC[i] << " ";
  }
  cout << "\n";
};

template <typename T>
void mat_print(vector<vector<T> > MAT){
  REP(i, MAT.size()){
    REP(j, MAT[i].size()){
      cout << MAT[i][j] << " ";
    }
    cout << "\n";
  }
};

template <typename CLASS1, typename CLASS2>
class HOGE{
  public:
    CLASS1 key;
    CLASS2 value;
    HOGE(void){
      return;
    };
    HOGE(CLASS1 key, CLASS2 value){
      this->key = key;
      this->value = value;
    };
    ~HOGE(void){
      return;
    };

    void print(void){
      cout << "key : " << key << ", value : " << value << "\n";
      return;
    };
    
    bool operator==(const HOGE &obj){
      return (this->value == obj.value);
    };
    bool operator<(const HOGE &obj){
      return (this->value < obj.value);
    };
    bool operator>(const HOGE &obj){
      return (this->value > obj.value);
    };
};

constexpr int INF = (1<<30);
constexpr ll INFLL = 1LL<<62;
constexpr long double EPS = 1e-12;
constexpr ll MOD = (ll)((1E+9)+7);

//using CHT_TYPE = ll;
using CHT_TYPE = long long;
//template <typename CHT_TYPE>
class ConvexHullTrickDynamic {
private:
	// 直線 **************************************************************
	struct Line {
		CHT_TYPE a, b; // y = ax + b
		mutable std::function<const Line*()> getSuc; // 次の直線へのポインタ (ソートで用いる)

		bool operator<(const Line& rhs) const {
			// 取得クエリでは次の直線との差分でソート
			if (rhs.b == IS_QUERY) {
				const Line* suc = getSuc();
				if (suc == nullptr) return false;
				const CHT_TYPE& x = rhs.a;
				return (suc->a - a)*x + suc->b - b > 0;
			}
			if (b == IS_QUERY) {
				const Line* suc = rhs.getSuc();
				if (suc == nullptr) return true;
				const CHT_TYPE& x = a;
				return (suc->a - rhs.a)*x + suc->b - rhs.b < 0;
			}

			// 通常の直線どうしは傾きソート
			return a < rhs.a;
		}
	};

	// 直線集合 **********************************************************
	class LinesSet : public std::multiset<Line> {
	private:
		// true -> 最小値クエリ, false -> 最大値クエリ
		bool flagMin;

	public:
		// コンストラクタ ( 第一引数falseで最大値クエリ,デフォルトで最小値クエリ )
		LinesSet(bool flagMin = true) : flagMin(flagMin) {};

		// 直線lが不必要であるかどうか
		inline bool isBad(iterator l) {
			const auto&& nel = std::next(l);
			if (l == begin()) { // lが傾き最小のとき
				if (nel == end()) return false; // lしかないなら必要
				return l->a == nel->a && l->b <= nel->b;
			}
			else {
				const auto&& prl = std::prev(l);
				if (nel == end()) return l->a == prl->a && l->b <= prl->b;
				return (prl->b - l->b) * (nel->a - l->a) >= (nel->b - l->b) * (prl->a - l->a);
			}
		}

		// 直線y=ax+bを追加する
		inline void add(CHT_TYPE a, CHT_TYPE b) {
			if (flagMin) a = -a, b = -b;
			auto&& it = insert({a, b});
			it->getSuc = [=] { return (std::next(it) == end() ? nullptr : &*std::next(it)); };
			if (isBad(it)) { erase(it); return; }
			while (std::next(it) != end() && isBad(std::next(it))) erase(std::next(it));
			while (it != begin() && isBad(std::prev(it))) erase(std::prev(it));
		}

		// 直線群の中でxの時に最小(最大)となる値を返す
		inline CHT_TYPE get(CHT_TYPE x) {
			auto&& l = *lower_bound(Line{x, IS_QUERY});
			if (flagMin) return -l.a * x - l.b;
			else return l.a * x + l.b;
		}
	};

	static const CHT_TYPE IS_QUERY = std::numeric_limits<CHT_TYPE>::lowest();
	LinesSet linesSet;

public:
	// コンストラクタ ( 第一引数falseで最大値クエリ,デフォルトで最小値クエリ )
	ConvexHullTrickDynamic(bool flagMin = true) : linesSet(flagMin) {}
	// 直線y=ax+bを追加する
	inline void add(CHT_TYPE a, CHT_TYPE b) { linesSet.add(a, b); }
	// あるxのときの直線集合での最小値を求める
	inline CHT_TYPE get(CHT_TYPE x) { return linesSet.get(x); }
};

int main(){
  cin.tie(0); // cut the cin and cout (default, std::flush is performed after std::cin)
  ios::sync_with_stdio(false); // cut the iostream and stdio (DON'T endl; BUT "\n";)

  ll N, C;
  cin >> N >> C;
  vll h(N);
  REP(i, N) cin >> h[i];

  vll dp(N, INFLL);

  ConvexHullTrickDynamic CHT;

  dp[0] = 0;
  CHT.add(-2*h[0], h[0]*h[0]);

  FOR(i, 1, N){
    dp[i] = CHT.get(h[i]) + h[i]*h[i]+C;
    CHT.add(-2*h[i], h[i]*h[i]+dp[i]);
  }

  cout << dp[N-1] << "\n";

  return 0;
}