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>
using namespace std;

struct line {
    long double k, b;
};

vector<line> vt;

long double intersect(line preLast, line last) {
    return - (double)(preLast.b - last.b) / (double)(preLast.k - last.k);
}

long double binarySearch(long double x) {
    int n = vt.size();
    int l = 1, r = n - 1, ans = -1;
    while (l <= r) {
        int m = (l + r) / 2;
        //cout << l << ' ' << m << ' ' << r << ' ' << ans << endl;
        line lcur = vt[m], lprev = vt[m - 1];
        long double xint = intersect(vt[m - 1], vt[m]);
        if (x >= xint) {
            ans = m, l = m + 1;
        } else {
            r = m - 1;
        }
    }
    //cout << vt[0].k * x + vt[0].b << endl;
    if (ans == -1)
        return vt[0].k * x + vt[0].b;
    else
        return vt[ans].k * x + vt[ans].b;
}

void addLine(line l) {
    while (vt.size() >= 2) {
        int n = vt.size();
        line preLast = vt[n - 2], last = vt[n - 1];
        //cout << "go " << preLast.k << ' ' << preLast.b << ' ' << last.k << ' ' << last.b << ' ' << l.k << ' ' << l.b << ' ' << intersect(preLast, last) << endl;
        //cout << - (preLast.b - l.b) * (preLast.k - last.k) << ' ' << - (preLast.b - last.b) * (preLast.k - l.k) << endl;
        long double x1 = intersect(preLast, last), x2 = intersect(preLast, l);
        if (x2 <= x1)
            vt.pop_back();
        else
            break;
    }
    vt.push_back(l);
}

void solve() {
    int n;
    long double c;
    cin >> n >> c;
    vector<long double> h(n);
    for (int i = 0; i < n; i++)
        cin >> h[i];
    vector<long double> dp(n);
    dp[0] = 0;
    addLine({-2 * h[0], h[0] * h[0]});
    for (int i = 1; i < n; i++) {
        dp[i] = binarySearch(h[i]) + h[i] * h[i] + c;
        //cout << binarySearch(h[i]) << ' ' << h[i] * h[i] << ' ' << c << ' ' << dp[i] << endl;
        addLine({-2 * h[i], dp[i] + h[i] * h[i]});
        //cout << vt.size() << endl;

        // cout << "line" << endl;
        // for (auto x : vt)
        //     cout << x.k << ' ' << x.b << endl;
        // cout << "endline" << endl;
    }
    cout << dp[n - 1];
}

int main() {
    ios::sync_with_stdio(NULL), cin.tie(0), cout.tie(0);
    cout.setf(ios::fixed), cout.precision(0);
    solve();
    return 0;
}