Cubes for Masha (思维+暴力)

构建数字游戏：最大整数挑战

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本文介绍了一个有趣的算法问题：使用一组立方体来构建从1到最大的整数。每个立方体的六面分别刻有不同的数字0到9。通过旋转立方体并排列它们，可以构建不同的数字。文章探讨了如何确定能够构建的最大整数，并提供了一个优化后的模拟算法实现。

Cubes for Masha

Absent-minded Masha got set of n cubes for her birthday.

At each of 6 faces of each cube, there is exactly one digit from 0 to 9. Masha became interested what is the largest natural x such she can make using her new cubes all integers from 1 to x.

To make a number Masha can rotate her cubes and put them in a row. After that, she looks at upper faces of cubes from left to right and reads the number.

The number can't contain leading zeros. It's not required to use all cubes to build a number.

Pay attention: Masha can't make digit 6 from digit 9 and vice-versa using cube rotations.

Input

In first line integer n is given (1 ≤ n ≤ 3) — the number of cubes, Masha got for her birthday.

Each of next n lines contains 6 integers a_i_j (0 ≤ a_i_j ≤ 9) — number on j-th face of i-th cube.

Output

Print single integer — maximum number x such Masha can make any integers from 1 to x using her cubes or 0 if Masha can't make even 1.

Examples

Input

3
0 1 2 3 4 5
6 7 8 9 0 1
2 3 4 5 6 7

Output

Input

3
0 1 3 5 6 8
1 2 4 5 7 8
2 3 4 6 7 9

Output

Note

In the first test case, Masha can build all numbers from 1 to 87, but she can't make 88 because there are no two cubes with digit 8.

优化暴力模拟，不可能出现比99大的数字，因为如果要出现比99大的数字，那么需要有1 1，2 2，3 3，4 4，5 5，6 6，7 7，8 8，9 9这样的话已经有18个数字，即使三个骰子也只是刚好18个数字。其中必不可少的是数字0，加上就是19。所以不可能出现比99大的数字。

code：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int main(){
    int n;
    int ans[4][7];
    int vis[100];
    memset(vis,0,sizeof(vis));
    scanf("%d",&n);
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= 6; j++){
            scanf("%d",&ans[i][j]);
        }
    }
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= 6; j++){
            vis[ans[i][j]] = 1;
            for(int k = 1; k <= n; k++){
                if(k == i) continue;
                for(int l = 1; l <= 6; l++){
                    vis[ans[i][j]*10+ans[k][l]] = 1;
                }
            }
        }
    }
    for(int i = 1; i <= 99; i++){
        if(!vis[i]){
            printf("%d\n",i-1);
            break;
        }
    }
    return 0;
}