Problem: You are given two bracket sequences (not necessarily regular) s and t consisting only of characters '(' and ')'. You want to construct the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous).

Recall what is the regular bracket sequence:

  * () is the regular bracket sequence; 
  * if S is the regular bracket sequence, then (S) is a regular bracket sequence; 
  * if S and T regular bracket sequences, then ST (concatenation of S and T) is a regular bracket sequence. 



Recall that the subsequence of the string s is such string t that can be obtained from s by removing some (possibly, zero) amount of characters. For example, "coder", "force", "cf" and "cores" are subsequences of "codeforces", but "fed" and "z" are not.

Input

The first line of the input contains one bracket sequence s consisting of no more than 200 characters '(' and ')'.

The second line of the input contains one bracket sequence t consisting of no more than 200 characters '(' and ')'.

Output

Print one line — the shortest regular bracket sequence that contains both given bracket sequences as subsequences (not necessarily contiguous). If there are several answers, you can print any.

Examples

Input


(())(()
()))()


Output


(())()()


Input


)
((


Output


(())


Input


)
)))


Output


((()))


Input


())
(()(()(()(


Output


(()()()(()()))

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize("O3", "unroll-loops")
#pragma GCC target("avx2")
using namespace std;
template <class T, class U>
inline void checkmin(T &x, U y) {
  if (y < x) x = y;
}
template <class T, class U>
inline void checkmax(T &x, U y) {
  if (y > x) x = y;
}
template <class T, class U>
inline bool ifmax(T &x, U y) {
  if (y > x) return x = y, true;
  return false;
}
template <class T, class U>
inline bool ifmin(T &x, U y) {
  if (y < x) return x = y, true;
  return false;
}
template <class T>
inline void sort(T &a) {
  sort(a.begin(), a.end());
}
template <class T>
inline void rsort(T &a) {
  sort(a.rbegin(), a.rend());
}
template <class T>
inline void reverse(T &a) {
  reverse(a.begin(), a.end());
}
template <class T, class U>
inline istream &operator>>(istream &str, pair<T, U> &p) {
  return str >> p.first >> p.second;
}
template <class T>
inline istream &operator>>(istream &str, vector<T> &a) {
  for (auto &i : a) str >> i;
  return str;
}
template <class T>
inline T sorted(T a) {
  sort(a);
  return a;
}
const int N = 205;
int dp[N][N][N];
struct pos {
  int p1 = -1, p2 = -1, balance;
  pos() {}
  pos(int p1, int p2, int balance) {
    this->p1 = p1;
    this->p2 = p2;
    this->balance = balance;
  }
};
pos p[N][N][N];
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(12);
  srand(94385);
  string second, t;
  cin >> second >> t;
  second = '0' + second;
  t = '0' + t;
  for (int i = 0; i < N; ++i)
    for (int j = 0; j < N; ++j)
      for (int k = 0; k < N; ++k) dp[i][j][k] = 1000000007;
  for (int balance = 0; balance < N; ++balance) dp[0][0][balance] = balance;
  for (int p1 = 0; p1 < signed(second.size()); ++p1)
    for (int p2 = 0; p2 < signed(t.size()); ++p2)
      if (p1 > 0 || p2 > 0) {
        for (int balance = 0; balance < N; ++balance) {
          pos &past = p[p1][p2][balance];
          if (balance > 0) {
            if (second[p1] == '(') {
              if (ifmin(dp[p1][p2][balance], dp[p1 - 1][p2][balance - 1] + 1))
                past = {p1 - 1, p2, balance - 1};
            }
            if (t[p2] == '(') {
              if (ifmin(dp[p1][p2][balance], dp[p1][p2 - 1][balance - 1] + 1))
                past = {p1, p2 - 1, balance - 1};
            }
            if (second[p1] == '(' && t[p2] == '(') {
              if (ifmin(dp[p1][p2][balance],
                        dp[p1 - 1][p2 - 1][balance - 1] + 1))
                past = {p1 - 1, p2 - 1, balance - 1};
            }
          }
          if (second[p1] == ')') {
            if (ifmin(dp[p1][p2][balance], dp[p1 - 1][p2][balance + 1] + 1))
              past = {p1 - 1, p2, balance + 1};
          }
          if (t[p2] == ')') {
            if (ifmin(dp[p1][p2][balance], dp[p1][p2 - 1][balance + 1] + 1))
              past = {p1, p2 - 1, balance + 1};
          }
          if (second[p1] == ')' && t[p2] == ')') {
            if (ifmin(dp[p1][p2][balance], dp[p1 - 1][p2 - 1][balance + 1] + 1))
              past = {p1 - 1, p2 - 1, balance + 1};
          }
        }
      }
  pair<int, pos> ans;
  ans.first = 1000000007;
  for (int balance = 0; balance < N; ++balance)
    if (ifmin(ans.first,
              dp[signed(second.size()) - 1][signed(t.size()) - 1][balance] +
                  balance))
      ans.second =
          pos(signed(second.size()) - 1, signed(t.size()) - 1, balance);
  pos curr = ans.second;
  string answer(curr.balance, ')');
  pos check = p[1][2][0];
  while (curr.p1 > 0 || curr.p2 > 0) {
    if (curr.balance > p[curr.p1][curr.p2][curr.balance].balance)
      answer = '(' + answer;
    else
      answer = ')' + answer;
    curr = p[curr.p1][curr.p2][curr.balance];
  }
  while (curr.balance) answer = '(' + answer, --curr.balance;
  cout << answer;
  return 0;
}