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 <deque>
#include <iostream>
using namespace std;

struct ConvexHullTrick {
    struct Line {
        long long a, b;
        long long get(long long x) { return a * x + b; }
    };
    deque<Line> que;
    bool check(Line a, Line b, Line c) {
        return (b.b - c.b) * (b.a - a.a) <= (a.b - b.b) * (c.a - b.a);
    }
    void add(long long a, long long b) {
        Line l{a, b};
        while(que.size() > 1 && check(que[que.size() - 2], que.back(), l))
            que.pop_back();
        que.push_back(l);
    }
    long long query(long long x) {
        if(que.empty()) return INT64_MAX;
        while(que.size() > 1 && que[0].get(x) >= que[1].get(x))
            que.pop_front();
        return que[0].get(x);
    }
};

int N;
long long C;
long long h[200009];

ConvexHullTrick cht;

long long dp[200009];

int main() {
    cin >> N >> C;
    for(int i = 0; i < N; i++)
        cin >> h[i];

    cht.add(-2 * h[0], h[0] * h[0]);
    for(int i = 1; i < N; i++) {
        dp[i] = cht.query(h[i]) + h[i] * h[i] + C;
        cht.add(-2 * h[i], dp[i] + h[i] * h[i]);
    }

    cout << dp[N - 1] << endl;

    return 0;
}