#使用动态规划取得记录最长连续子串的二维数组，再从二维数组中获取需要的子串在原字符串中的位置

    try:
        c=input()
        d=input()
    except:
        break
    else:
        if len(c) > len(d):
            a = d
            b = c
        else:
            a = c
            b = d
        len_a = len(a)+1
        len_b = len(b)+1
        dp = [[0 for i in range(len(a) + 1)] for j in range(len(b) + 1)]
        for j in range(1,len_b):
            for i in range(1,len_a):
                if a[i-1] == b[j-1]:
                    dp[j][i] = dp[j-1][i-1] + 1
                else:
                    dp[j][i] = 0
        maxx=0
        maxx_str = ''
        indexx = ()
        for i in range(1,len_a):
            for j in range(1,len_b):
                if dp[j][i] >maxx:
                    maxx = dp[j][i]
                    maxx_str = "".join(a[dp[j].index(maxx)-maxx:dp[j].index(maxx)])
                    indexx = (dp[j].index(maxx)-maxx,dp[j].index(maxx))
#         print(dp)
#         print(a,indexx)
        print(maxx_str)