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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

const int mxN = 2e5 + 5;
int N;
ll h[mxN], C, dp[mxN];

struct line {
    ll m, c;

    line(ll _m, ll _c) {
        m = _m;
        c = _c;
    }
    ll value(ll x) {
        return (m * x + c);
    }
};

deque<line> lines;

inline ll square(ll x) {
    return (x * x);
}

bool chk(line& l1, line& l2, line& l3) {
    return (l3.c - l2.c) * (l2.m - l1.m) >= (l2.c - l1.c) * (l3.m - l2.m);
}

void add_line(ll _m, ll _c) {
    line new_line = line(_m, _c);
    while (lines.size() >= 2 && chk(lines[(int) lines.size() - 2], lines[(int) lines.size() - 1], new_line)) {
        lines.pop_back();
    }
    lines.emplace_back(new_line);
}

ll query(ll x) {
    while (lines.size() > 1 && lines[0].value(x) > lines[1].value(x)) {
        lines.pop_front();
    }
    return lines[0].value(x);
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    cin >> N >> C;
    for (int i = 0; i < N; i++) {
        cin >> h[i];
    }
    dp[0] = 0;
    add_line(-2 * h[0], square(h[0]) + dp[0]);
    for (int i = 1; i < N; i++) {
        ll val = query(h[i]);
        dp[i] = val + square(h[i]) + C;
        add_line(-2 * h[i], square(h[i]) + dp[i]);
    }
    cout << dp[N - 1];

    return 0;
}