//
// Monge 単一始点最短路
//   頂点数 N+1 の DAG, 頂点 i, j 間のコスト f(i, j) が Monge であることを仮定
//   O(N log N)
//
// verified
//   AtCoder EDPC Z - Frog 3
//     https://atcoder.jp/contests/dp/tasks/dp_z 
//
//   Codeforces Round 189 (Div. 1) C. Kalila and Dimna in the Logging Industry
//     https://codeforces.com/contest/319/problem/C 
// 
//   yukicoder No.705 ゴミ拾い Hard
//     https://yukicoder.me/problems/no/705 
//
// Reference:
//   noshi: 簡易版 LARSCH Algorithm
//     https://noshi91.hatenablog.com/entry/2023/02/18/005856
//  


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


//------------------------------//
// Utility
//------------------------------//

using ll = long long;
using i128 = __int128_t;
using u128 = __uint128_t;
using pint = pair<int, int>;
using pll = pair<long long, long long>;
using tll = array<long long, 3>;
using fll = array<long long, 4>;
using vint = vector<int>;
using vll = vector<long long>;
using dint = deque<int>;
using dll = deque<long long>;
using vvint = vector<vector<int>>;
using vvll = vector<vector<long long>>;
using vpll = vector<pair<long long, long long>>;
template<class T> using min_priority_queue = priority_queue<T, vector<T>, greater<T>>;

template<class S, class T> inline bool chmax(S &a, T b) { return (a < b ? a = b, 1 : 0); }
template<class S, class T> inline bool chmin(S &a, T b) { return (a > b ? a = b, 1 : 0); }
template<class S, class T> inline auto maxll(S a, T b) { return max(ll(a), ll(b)); }
template<class S, class T> inline auto minll(S a, T b) { return min(ll(a), ll(b)); }
template<class T> auto max(const T &a) { return *max_element(a.begin(), a.end()); }
template<class T> auto min(const T &a) { return *min_element(a.begin(), a.end()); }
template<class T> auto argmax(const T &a) { return max_element(a.begin(), a.end()) - a.begin(); }
template<class T> auto argmin(const T &a) { return min_element(a.begin(), a.end()) - a.begin(); }
template<class T> auto accum(const vector<T> &a) { return accumulate(a.begin(), a.end(), T()); }
template<class T> auto accum(const deque<T> &a) { return accumulate(a.begin(), a.end(), T()); }

#define REP(i, a) for (long long i = 0; i < (long long)(a); i++)
#define REP2(i, a, b) for (long long i = a; i < (long long)(b); i++)
#define RREP(i, a) for (long long i = (a)-1; i >= (long long)(0); --i)
#define RREP2(i, a, b) for (long long i = (b)-1; i >= (long long)(a); --i)
#define EB emplace_back
#define PF push_front
#define PB push_back
#define MP make_pair
#define FI first
#define SE second
#define ALL(x) x.begin(), x.end()
#define COUT(x) cout << #x << " = " << (x) << " (L" << __LINE__ << ")" << endl

// input
template<class T> istream& operator >> (istream &is, vector<T> &P)
{ for (int i = 0; i < (int)P.size(); ++i) cin >> P[i]; return is; }
template<class T> istream& operator >> (istream &is, deque<T> &P)
{ for (int i = 0; i < (int)P.size(); ++i) cin >> P[i]; return is; }
template<class T> istream& operator >> (istream &is, vector<vector<T>> &P)
{ for (int i = 0; i < (int)P.size(); ++i) cin >> P[i]; return is; }

// output
template<class S, class T> ostream& operator << (ostream &s, const pair<S, T> &P)
{ return s << '<' << P.first << ", " << P.second << '>'; }
template<class T> ostream& operator << (ostream &s, const array<T, 2> &P)
{ return s << '<' << P[0] << "," << P[1] << '>'; }
template<class T> ostream& operator << (ostream &s, const array<T, 3> &P)
{ return s << '<' << P[0] << "," << P[1] << "," << P[2] << '>'; }
template<class T> ostream& operator << (ostream &s, const array<T, 4> &P)
{ return s << '<' << P[0] << "," << P[1] << "," << P[2] << "," << P[3] << '>'; }
template<class T> ostream& operator << (ostream &s, const vector<T> &P)
{ for (int i = 0; i < P.size(); ++i) { if (i > 0) { s << " "; } s << P[i]; } return s; }
template<class T> ostream& operator << (ostream &s, const deque<T> &P)
{ for (int i = 0; i < P.size(); ++i) { if (i > 0) { s << " "; } s << P[i]; } return s; }
template<class T> ostream& operator << (ostream &s, const vector<vector<T>> &P)
{ for (int i = 0; i < P.size(); ++i) { s << endl << P[i]; } return s << endl; }
template<class T> ostream& operator << (ostream &s, const set<T> &P)
{ for (auto it : P) { s << "<" << it << "> "; } return s; }
template<class T> ostream& operator << (ostream &s, const multiset<T> &P)
{ for (auto it : P) { s << "<" << it << "> "; } return s; }
template<class T> ostream& operator << (ostream &s, const unordered_set<T> &P)
{ for (auto it : P) { s << "<" << it << "> "; } return s; }
template<class S, class T> ostream& operator << (ostream &s, const map<S, T> &P)
{ for (auto it : P) { s << "<" << it.first << "->" << it.second << "> "; } return s; }
template<class S, class T> ostream& operator << (ostream &s, const unordered_map<S, T> &P)
{ for (auto it : P) { s << "<" << it.first << "->" << it.second << "> "; } return s; }
void yes(bool a) { cout << (a ? "yes" : "no") << endl; }
void YES(bool a) { cout << (a ? "YES" : "NO") << endl; }
void Yes(bool a) { cout << (a ? "Yes" : "No") << endl; }
const vector<int> DX = {1, 0, -1, 0, 1, -1, 1, -1};
const vector<int> DY = {0, 1, 0, -1, 1, -1, -1, 1};






// noshi's simplified LARSCH
// find shortest path from vertex 0 on DAG with monotone cost in O(N log N)
// vertex: 0, 1, 2, ..., N, f(i, j) must be Monge
template<class VAL> struct MongeShortestPath {
    VAL INF = numeric_limits<VAL>::max() / 2;
    int CNT_INF = numeric_limits<int>::max() / 2;

    // results
    vector<VAL> dp;
    vector<int> cnt, prev;

    // solver
    template<class FUNC> vector<pair<VAL, int>> solve(int N, const FUNC &f, bool minimize_cnt = true) {
        dp.assign(N + 1, INF);
        cnt.assign(N + 1, CNT_INF);
        prev.assign(N + 1, 0);
        dp[0] = 0, cnt[0] = 0;

        auto relax = [&](int l, int r) -> void {
            VAL val = dp[l] + f(l, r);
            int c = cnt[l] + 1;
            if (dp[r] > val || (dp[r] == val && (minimize_cnt ? c < cnt[r] : c > cnt[r]))) {
                dp[r] = val;
                cnt[r] = c;
                prev[r] = l;
            }
        };
        auto rec = [&](auto &&rec, int l, int r) -> void {
            if (r - l <= 1) return;
            int m = (l + r) / 2;
            for (int k = prev[l]; k <= prev[r]; k++) relax(k, m);
            rec(rec, l, m);
            for (int k = l + 1; k <= m; k++) relax(k, r);
            rec(rec, m, r);
        };

        if (N > 0) relax(0, N), rec(rec, 0, N);
        vector<pair<VAL, int>> res(N + 1, make_pair(numeric_limits<VAL>::max() / 2, -1));
        res[0].first = VAL(0);
        for (int i = 1; i <= N; i++) res[i] = {dp[i], prev[i]};
        return res;
    }

    vector<int> reconstruct() {
        int N = (int)dp.size() - 1;
        vector<int> path;
        for (int v = N; v > 0; v = prev[v]) path.emplace_back(v);
        path.emplace_back(0);
        reverse(path.begin(), path.end());
        return path;
    }
};


//------------------------------//
// Examples
//------------------------------//

// AtCoder EDPC Z - Frog 3
/*
    H は単調増加数列
    chmin(dp[j], dp[i] + (H[j] - H[i])^2 + C)
    i -> j のコスト：(H[j] - H[i])^2 ...... 差の凸関数は Monge
    スタート: 0, ゴール: N-1
*/
void EDPC_Z() {
    long long N, C;
    cin >> N >> C;
    vector<long long> H(N);
    for (long long i = 0; i < N; i++) cin >> H[i];
    auto func = [&](int i, int j) -> long long {
        return (H[j] - H[i]) * (H[j] - H[i]) + C;
    };
    MongeShortestPath<long long> msp;
    auto res = msp.solve(N-1, func);
    cout << res[N-1].first << endl;
}

// Codeforces Round 189 (Div. 1) C. Kalila and Dimna in the Logging Industry
/*
    A: 単調増加, B: 単調減少, ともに長さ N
    i -> j のコストが、B[i] × A[j] で与えられる ..... 単調増加 × 単調減少は Monge
    スタート: 0, ゴール: N-1
*/
void Codeforces_189_C() {
    long long N;
    cin >> N;
    vector<long long> A(N), B(N);
    for (int i = 0; i < N; i++) cin >> A[i];
    for (int i = 0; i < N; i++) cin >> B[i];
    auto func = [&](int i, int j) -> long long {
        return B[i] * A[j];
    };
    MongeShortestPath<long long> msp;
    auto res = msp.solve(N-1, func);
    cout << res[N-1].first << endl;
}

// yukicoder No.705 ゴミ拾い Hard
/*
    A, X, Y: N 個
    これらを区間に分割していく
    　dp[j] = min_{0 ≦ i < j}(dp[i] + |A[j-1] - X[i]|^3 + |-Y[i]|^3)
    i -> j のコスト：|A[j-1] - X[i]|^3 + |-Y[i]|^3 ...... 差の凸関数 (Monge) + 縞々 (Monge) -> Monge
    スタート: 0, ゴール: N
*/
void yukicoder_705() {
    int N;
    cin >> N;
    vector<long long> A(N), X(N), Y(N);
    for (int i = 0; i < N; i++) cin >> A[i];
    for (int i = 0; i < N; i++) cin >> X[i];
    for (int i = 0; i < N; i++) cin >> Y[i];
    auto func = [&](int i, int j) -> long long {
        long long dx = abs(A[j-1] - X[i]), dy = abs(Y[i]);
        return dx * dx * dx + dy * dy * dy;
    };
    MongeShortestPath<long long> msp;
    auto res = msp.solve(N, func);
    cout << res[N].first << endl;
}


int main() {
    //EDPC_Z();
    //Codeforces_189_C();
    yukicoder_705();
}