Problem:
Score : 100 points

Problem Statement

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

Notes

A subsequence of a string x is the string obtained by removing zero or more characters from x and concatenating the remaining characters without changing the order.

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.

Sample Input 1

axyb
abyxb

Sample Output 1

axb

The answer is axb or ayb; either will be accepted.

Sample Input 2

aa
xayaz

Sample Output 2

aa

Sample Input 3

a
z

Sample Output 3

The answer is (an empty string).

Sample Input 4

abracadabra
avadakedavra

Sample Output 4

aaadara

Solution in Pascal (FPC 2.6.2) (judge verdict: Accepted)

uses
  math;
var
  f:array[0..3033,0..3033] of longint;
  n,m,j,i:longint;
  s,r,kq:ansistring;
BEGIN
  ///assign(input,'a.in'); reset(input);
  readln(s);  n := length(s);
  readln(r);  m := length(r);
  
  fillchar(f,sizeof(f),0);
  for i:=1 to n do
    for j:=1 to m do
      if s[i] <> r[j] then f[i,j] := max(f[i-1,j], f[i,j-1])
      else f[i,j] := 1 + f[i-1,j-1];
  kq := '';
  i:=n; j:=m;
  while((i>0) and (j>0)) do begin
    if (f[i,j] = f[i-1,j-1] + 1) and (s[i]=r[j]) then begin kq:=s[i] + kq; dec(i); dec(j); end
    else
      if f[i,j] = f[i-1,j] then dec(i) else dec(j);
  end;
  writeln(kq);
END.