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 PyPy3 (2.4.0) (judge verdict: Time Limit Exceeded)

S = input()
T = input()
LS = len(S); LT = len(T)

dp = [[-1]*(LT+1) for i in range(LS+1)]
nxt = [[(0, None, None)]*(LT+1) for i in range(LS+1)]
P = [[None]*(LT+1) for i in range(LS+1)]
Q = [[None]*(LT+1) for i in range(LS+1)]
R = [[0]*(LT+1) for i in range(LS+1)]
for i in range(LS-1, -1, -1):
    for j in range(LT-1, -1, -1):
        r0 = dp[i+1][j]; r1 = dp[i][j+1]
        r = max(r0, r1)
        if r0 < r1:
            rr = 0; p0 = i; q0 = j+1
        else:
            rr = 0; p0 = i+1; q0 = j
        if S[i] == T[j]:
            r2 = dp[i+1][j+1] + 1
            if r < r2:
                r = r2
                rr = 1; p0 = i+1; q0 = j+1
        dp[i][j] = r
        R[i][j] = rr
        P[i][j] = p0; Q[i][j] = q0

i = j = 0
ans = []
while i is not None:
    if R[i][j]:
        ans.append(S[i])
    i, j = P[i][j], Q[i][j]
print(*ans, sep='')