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++14 (GCC 5.4.1) (judge verdict: Runtime Error)

//
// Created by Hideaki Imamura on 2020-03-15.
//
# include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef pair <ll, ll> l_l;
typedef pair<int, int> i_i;

template<class T>
inline bool chmax(T &a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

template<class T>
inline bool chmin(T &a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}

# define EPS (1e-7)
# define INF (1e9)
# define PI (acos(-1))
//const ll mod = 1000000007;

string s, t;

int main() {
    cin >> s >> t;

    int n = s.length();
    int m = t.length();
    vector<vector<pair<int, string>>> dp(n + 1, vector<pair<int, string>>(m + 1, make_pair(0, "")));
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            if (s[i-1] == t[j-1]) {
                dp[i][j] = make_pair(dp[i-1][j-1].first + 1, dp[i-1][j-1].second + s[i-1]);
            } else {
                dp[i][j] = dp[i-1][j].first >= dp[i][j-1].first ? dp[i-1][j] : dp[i][j-1];
            }
        }
    }
    cout << dp[n][m].second << endl;
    return 0;
}