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 Kotlin (1.0.0) (judge verdict: Time Limit Exceeded)

fun main(args: Array<String>) {
    val s = readLine()!!
    val t = readLine()!!

    val dp = Array(s.length + 1) { Array(t.length + 1) { 0 } }

    for (i in 0 until s.length) {
        for (j in 0 until t.length) {
            if (s[i] == t[j]) dp[i + 1][j + 1] = Math.max(dp[i + 1][j + 1], dp[i][j] + 1)
            dp[i + 1][j + 1] = Math.max(dp[i + 1][j + 1], dp[i + 1][j])
            dp[i + 1][j + 1] = Math.max(dp[i + 1][j + 1], dp[i][j + 1])
        }
    }

    var res = ""
    var i = s.length
    var j = t.length
    while (i > 0 && j > 0) {

        if (dp[i][j] == dp[i - 1][j]) {
            i--
            continue
        }

        if (dp[i][j] == dp[i][j - 1]) {
            j--
            continue
        }

        res = s[i - 1] + res
        i--
        j--
    }

    println(res)
}