#include <iostream>
#include <algorithm>
#include <iomanip>
#include <vector>
#include <queue>
#include <deque>
#include <set>
#include <map>
#include <tuple>
#include <cmath>
#include <numeric>
#include <functional>
#include <cassert>

#define debug_value(x) cerr << "line" << __LINE__ << ":<" << __func__ << ">:" << #x << "=" << x << endl;
#define debug(x) cerr << "line" << __LINE__ << ":<" << __func__ << ">:" << x << endl;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

using namespace std;
typedef long long ll;

template<typename T>
vector<vector<T>> vec2d(int n, int m, T v){
    return vector<vector<T>>(n, vector<T>(m, v));
}

template<typename T>
vector<vector<vector<T>>> vec3d(int n, int m, int k, T v){
    return vector<vector<vector<T>>>(n, vector<vector<T>>(m, vector<T>(k, v)));
}

template<typename T>
void print_vector(vector<T> v, char delimiter=' '){
    if(v.empty()) {
        cout << endl;
        return;
    }
    for(int i = 0; i+1 < v.size(); i++) cout << v[i] << delimiter;
    cout << v.back() << endl;
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout << setprecision(10) << fixed;
    int n; cin >> n;
    auto dp = vec3d(n+1, 1<<6, 5, 0ll);
    dp[0][0][0] = 1;
    for(int i = 0; i < n; i++){
        vector<ll> v(6);
        for(int j = 0; j < 6; j++) cin >> v[j];
        vector<ll> p(1<<6, 1ll);
        for(int s = 0; s < (1<<6); s++){
            for(int j = 0; j < 6; j++){
                if(s&(1<<j)) p[s] *= v[j];
            }
        }
        for(int s = 0; s < (1<<6); s++){
            for(int k = 0; k <= 4; k++) chmax(dp[i+1][s][k], dp[i][s][k]);
            for(int t = 0; t < (1<<6); t++){
                if((s|t) > t) continue;
                int u = t-s;
                for(int k = 0; k <= 4; k++){
                    if(k <= 3) chmax(dp[i+1][t][k+1], dp[i][s][k]*p[u]);
                }
            }
        }
    }
    cout << dp[n][(1<<6)-1][4] << endl;
}