//思路：动态规划经典问题，加一个start标记即可,注意将较短子串最先出现的那个输出

#include<iostream>

#include<vector>

#include<string>

using namespace std;

void findMaxCommonStr(string s1,string s2)

{    

    if(s1.length()>s2.length())            

        swap(s1,s2);//s1用于保存较短的子串    

    int len1=s1.length(),len2=s2.length();    

    int maxLen=,start=;    

    vector<vector<int> >dp(len1+,vector<int>(len2+,));    

    for(int i=;i<=len1;++i)        

        for(int j=;j<=len2;++j)        

        {            

            if(s1[i-]==s2[j-])           

            {                

                dp[i][j]=dp[i-][j-]+;                

                if(dp[i][j]>maxLen)                

                {                    

                    maxLen=dp[i][j];                    

                    start=i-maxLen;//记录最长公共子串的起始位置                

                }            

            }       

        }   

    cout<<s1.substr(start,maxLen)<<endl;

}

int main()

{   

    string s1,s2;   

    while(cin>>s1>>s2)   

    {       

        findMaxCommonStr(s1,s2);  

    }   

    return ;

}