B - Cubes for Masha /* Codeforces Round #444 (div.2) */

B. Cubes for Masha

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

    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 aij (0 ≤ aij ≤ 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

87

input

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

output

98

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.

 

题意：

给你n行数字，每行6个0~9的数字，每行的数字最多取一个或者不取，找到最小的不能组成的数字x ，且1~x-1都能组成。

 

代码：

#include <iostream>
#include <map>
#include <cstdio>
#include <vector>
using namespace std;
int n,t;
int a[20];
int hh[3]; ///统计对于一个数字而言，是否已经用过了每一行

map<int,vector<int> >mp;
int temp;
void solve(int ans)
{
    ///判断行的使用情况，使用超过一次以上就说明不行
    for(int i=0;i<n;i++)
        if(hh[i]>1) return ; 
    
    ///数字组成完毕
    if(ans==0) {temp=1;return;}
    ///缺少组成这一部分的数
    if(mp[ans%10].size()==0) return;

    else
    {
        for(int i=0;i<mp[ans%10].size();i++)
        {
            hh[mp[ans%10][i]]++;
            solve(ans/10);
            hh[mp[ans%10][i]]--;
            ///把每一种都试
        }
    }
}

int main()
{
    cin>>n;
    for(int i=0;i<n;i++)  ///输入，mp里存放每一个数字出现的行数
       for(int j=0;j<6;j++)
          {
              cin>>a[t++];
              mp[a[t-1]].push_back(i);
          }
          
    temp=1;
    int ans=1; ///递增遍历数字，直到找到不能组成的数字为止
    while(temp){
        hh[0]=hh[1]=hh[2]=temp=0; ///清零
        solve(ans);
        ans++;
    }
    ///输出那个不能组成的最小的数
    cout<<ans-2<<endl;
    return 0;
}