Problem: You are given strings s and t. Find one longest string that is a subsequence of both s and t.

Constraints

* s and t are strings consisting of lowercase English letters.
* 1 \leq |s|, |t| \leq 3000

Input

Input is given from Standard Input in the following format:


s
t


Output

Print one longest string that is a subsequence of both s and t. If there are multiple such strings, any of them will be accepted.

Examples

Input

axyb
abyxb


Output

axb


Input

aa
xayaz


Output

aa


Input

a
z


Output




Input

abracadabra
avadakedavra


Output

aaadara

Solution (in JAVA):

Solution:
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String s = sc.next();
        String t = sc.next();
        // dp[i][j]：ｓのｉ文字目までと、ｔのｊ文字目までの最長共通部分列の長さ
        int[][] dp = new int[s.length()+1][t.length()+1];
        for(int i = 0; i < s.length(); i++){
            for(int j = 0; j < t.length(); j++){
                if(s.charAt(i) == t.charAt(j)) dp[i+1][j+1] = dp[i][j] + 1;
                else dp[i+1][j+1] = Math.max(dp[i+1][j], dp[i][j+1]);
            }
        }
        StringBuilder sb = new StringBuilder();
        // LCSを復元する
        for(int i = s.length(), j = t.length(); i > 0 && j > 0;){
            if(s.charAt(i-1) == t.charAt(j-1)){
                sb.append(s.charAt(i-1));
                i--; j--;
            }else if(dp[i][j] == dp[i-1][j]) i--;
              else j--;
        }
        System.out.println(sb.reverse().toString());
        sc.close();
    }

}