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: Runtime Error)

#include <bits/stdc++.h>
#define int long long
#define lint long long
#define pii pair<int,int>
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n) FOR(i,0,n)
#define ALL(c) (c).begin(),(c).end()
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
#define MINF(a) memset(a,0x3f,sizeof(a))
#define POW(n) (1LL<<(n))
#define POPCNT(n) (__builtin_popcount(n))
#define IN(i,a,b) (a <= i && i <= b)
using namespace std;
template <typename T> inline bool CHMIN(T& a,T b) { if(a>b) { a=b; return 1; } return 0; }
template <typename T> inline bool CHMAX(T& a,T b) { if(a<b) { a=b; return 1; } return 0; }
template <typename T> inline void SORT(T& a) { sort(ALL(a)); }
template <typename T> inline void REV(T& a) { reverse(ALL(a)); }
template <typename T> inline void UNI(T& a) { sort(ALL(a)); a.erase(unique(ALL(a)),a.end()); }
template <typename T> inline T LB(vector<T>& v, T a) { return *lower_bound(ALL(v),a); }
template <typename T> inline int LBP(vector<T>& v, T a) { return lower_bound(ALL(v),a) - v.begin(); }
template <typename T> inline T UB(vector<T>& v, T a) { return *upper_bound(ALL(v),a); }
template <typename T> inline int UBP(vector<T>& v, T a) { return upper_bound(ALL(v),a) - v.begin(); }
template <typename T1, typename T2> ostream& operator<< (ostream& os, const pair<T1,T2>& p) { os << p.first << " " << p.second; return os; }
template <typename T1, typename T2> istream& operator>> (istream& is, pair<T1,T2>& p) { is >> p.first >> p.second; return is; }
template <typename T> ostream& operator<< (ostream& os, const vector<T>& v) { REP(i,v.size()) { if (i) os << " "; os << v[i]; } return os; }
template <typename T> istream& operator>> (istream& is, vector<T>& v) { for(T& in : v) is >> in; return is; }
template <typename T = int> vector<T> make_v(size_t a) { return vector<T>(a); }
template <typename T, typename... Ts> auto make_v(size_t a, Ts... ts) { return vector<decltype(make_v<T>(ts...))>(a,make_v<T>(ts...)); }
template <typename T, typename V> typename enable_if<is_class<T>::value == 0>::type fill_v(T &t, const V &v) { t = v; }
template <typename T, typename V> typename enable_if<is_class<T>::value != 0>::type fill_v(T &t, const V &v) { for(auto &e : t) fill_v(e,v); }
const lint MOD = 1000000007;
const lint INF = 0x3f3f3f3f3f3f3f3f;
const double EPS = 1e-10;

struct LiChaoTree {
    int n;
    vector<int> xs;  // 座標
    vector<int> ps;  // 直線の傾き
    vector<int> qs;  // 直線の切片

    LiChaoTree(vector<int> &xin)
    {
        for (n = 1; n < xin.size(); n <<= 1);
        ps.assign(n << 1, 0);
        qs.assign(n << 1, INF);
        xs.resize(n, *max_element(xin.begin(), xin.end()) + 1);
        for (int i = 0; i < xin.size(); ++i) {
            xs[i] = xin[i];
        }
    }

    int f(int p, int q, int x)
    {
        return p * x + q;
    }

    // 追加処理
    // [l, r)
    void add(int p, int q, int k, int l, int r) {
        int m = (l + r) / 2;
        if (ps[k] == 0 && qs[k] == INF) {
            ps[k] = p;
            qs[k] = q;
            return;
        }
        bool left = (f(p, q, xs[l]) < f(ps[k], qs[k], xs[l]));
        bool mid = (f(p, q, xs[m]) < f(ps[k], qs[k], xs[m]));
        bool right = (f(p, q, xs[r - 1]) < f(ps[k], qs[k], xs[r - 1]));
        if (left && right) {
            ps[k] = p;
            qs[k] = q;
            return;
        }
        if (!left && !right)
            return;
        if (mid) {
            swap(ps[k], p);
            swap(qs[k], q);
        }
        if (left != mid) {
            add(p, q, (k << 1) | 0, l, m);
        } else {
            add(p, q, (k << 1) | 1, m, r);
        }
    }

    // 直線追加
    // O(log n)
    void add(int p, int q) {
        return add(p, q, 1, 0, n);
    }

    // 線分追加 [a, b)
    void add(int p, int q, int a, int b) {
        int L = a + n, R = b + n;
        int a0 = a, b0 = b;
        int sz = 1;
        while (L < R) {
            if (L & 1) {
                add(p, q, L, a0, a0 + sz);
                L++;
                a0 += sz;
            }
            if (R & 1) {
                R--;
                b0 -= sz;
                add(p, q, R, b0, b0 + sz);
            }
            L >>= 1; R >>= 1;
            sz <<= 1;
        }
    }

    // k番目の座標の最小値 (0-indexed)
    // O(log n)
    int query(int k) {
        int x = xs[k];
        k += n;
        int res = INF;
        while (k > 0) {
            res = min(res, f(ps[k], qs[k], x));
            k >>= 1;
        }
        return res;
    }

    void print()
    {
        int lev = 0;
        while ((1 << lev) < n) lev++;
        cout << "(傾き, 切片):" << endl;
        for (int i = 0; i <= lev; ++i) {
            for (int j = 0; j < (1 << i); ++j) {
                cout << " (" << ps[(1 << i) + j] << ", " << qs[(1 << i) + j] << ")";
            }
            cout << endl;
        }
        cout << endl;
    }
};

void _main() {
    int N, C;
    cin >> N >> C;
    vector<int> h(N);
    for (int i = 0; i < N; ++i) {
        cin >> h[i];
    }
    vector<int> dp(N, 0);
    LiChaoTree lct(h);
    for (int i = 0; i < N; ++i) {
        lct.add(-2 * h[i], h[i] * h[i] + dp[i] + C);
        dp[i + 1] = lct.query(i + 1) + h[i + 1] * h[i + 1];
    }
    cout << dp[N - 1] << endl;
}

signed main(signed argc, char **argv) {
    if (argc > 1) {
        if (strchr(argv[1], 'i'))
            freopen("input.txt", "r", stdin);
        if (strchr(argv[1], 'o'))
            freopen("output.txt", "w", stdout);
    }
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(10);
    _main();
    return 0;
}