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 C (GCC 5.4.1) (judge verdict: Runtime Error)

#include<stdio.h>
#include<string.h>
#define LEN 3000

int max(int a, int b){return (a>b)?a:b;}

int lcs[LEN][LEN]={0};
char word[LEN]="";

int main(void){
	int i;
	int j;
	int k;
	int s_len;
	int t_len;
	char s[LEN];
	char t[LEN];
	scanf("%s%s",s,t);
	s_len=strlen(s);
	t_len=strlen(t);
	for(i=0;i<s_len+1;i++){
		lcs[i][0]=0;
	}
	for(i=0;i<t_len+1;i++){
		lcs[0][i]=0;
	}
	for(i=1;i<s_len+1;i++){
		for(j=1;j<t_len+1;j++){
			if(s[i-1]==t[j-1]){
				lcs[i][j]=lcs[i-1][j-1]+1;
			}else{
				lcs[i][j]=max(lcs[i-1][j],lcs[i][j-1]);
			}
		}
	}
	i=s_len;
	j=t_len;
	k=1;
	while((i>0)&&(j>0)){
		if(lcs[i][j]==lcs[i][j-1]){
			j--;
		}else if(lcs[i][j]==lcs[i-1][j]){
			i--;
		}else{
			word[lcs[s_len][t_len]-k]=s[i-1];
			i--;
			j--;
			k++;
		}
	}
	printf("%s",word);
	return 0;
}