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>
bool local = false;
using namespace std;
template <class T>
string toString(T x) {
  ostringstream sout;
  sout << x;
  return sout.str();
}
template <class T>
void print(T v) {
  cout << ((int)(v).size()) << "\n";
  for (auto x : v) cout << x << ' ';
  cout << "\n";
};
template <class T>
void print1(T v) {
  cout << ((int)(v).size()) << "\n";
  for (auto x : v) cout << x + 1 << ' ';
  cout << "\n";
};
template <class T1, class T2>
ostream &operator<<(ostream &o, pair<T1, T2> x) {
  return o << x.first << ' ' << x.second;
}
template <class T1, class T2>
istream &operator>>(istream &o, pair<T1, T2> &x) {
  return o >> x.first >> x.second;
}
template <class T1, class T2>
pair<T1, T2> operator+(pair<T1, T2> a, pair<T1, T2> b) {
  a.first += b.first;
  a.second += b.second;
  return a;
}
template <class T1, class T2>
pair<T1, T2> operator-(pair<T1, T2> a, pair<T1, T2> b) {
  a.first -= b.first;
  a.second -= b.second;
  return a;
}
template <class T1, class T2>
void operator+=(pair<T1, T2> &a, pair<T1, T2> b) {
  a.first += b.first;
  a.second += b.second;
}
template <class T1, class T2>
void operator-=(pair<T1, T2> &a, pair<T1, T2> b) {
  a.first -= b.first;
  a.second -= b.second;
}
int nint() {
  int x;
  cin >> x;
  return x;
}
double getTime() { return clock() / (double)CLOCKS_PER_SEC; };
mt19937 rnd(0);
int rand(int n) { return rnd() % n; }
int rand(int l, int r) { return rnd() % (r - l + 1) + l; }
const int mod = 1000000000 + 7;
void initIO() {
  if (local) {
    freopen("input.txt", "r", stdin);
    srand((int)time(0));
    rnd.seed((int)time(0));
  } else {
    {
      ios::sync_with_stdio(0);
      cin.tie(0);
      cout.tie(0);
    };
  }
}
void solve();
int main() {
  initIO();
  int tc = 1;
  for (int tt = 0; tt < (tc); tt++) {
    solve();
  }
  if (local)
    cout << "\n"
         << "time = " << getTime() << "\n";
}
int n;
int m;
int table[210][210][420];
int ne[210][210][420];
string s1, s2;
int getans(int p1, int p2, int b) {
  if (b < 0 || b > n + m) return mod;
  if (p1 == n && p2 == m && b == 0) return 0;
  int &ans = table[p1][p2][b];
  if (ans != -1) return ans;
  ans = getans(min(n, p1 + (s1[p1] == '(')), min(m, p2 + (s2[p2] == '(')),
               b + 1) +
        1;
  ne[p1][p2][b] = 1;
  if (b) {
    int t = getans(min(n, p1 + (s1[p1] == ')')), min(m, p2 + (s2[p2] == ')')),
                   b - 1) +
            1;
    if (t < ans) {
      ans = t;
      ne[p1][p2][b] = 2;
    }
  }
  return ans;
}
void solve() {
  cin >> s1 >> s2;
  n = ((int)(s1).size());
  m = ((int)(s2).size());
  s1 += " ";
  s2 += " ";
  memset(table, -1, sizeof(table));
  getans(0, 0, 0);
  int p1 = 0, p2 = 0, b = 0;
  while (p1 < n || p2 < m || b) {
    int o = ne[p1][p2][b];
    if (o == 1) {
      cout << "(";
      if (p1 < n && s1[p1] == '(') p1++;
      if (p2 < m && s2[p2] == '(') p2++;
      b++;
    } else {
      cout << ")";
      if (p1 < n && s1[p1] == ')') p1++;
      if (p2 < m && s2[p2] == ')') p2++;
      b--;
    }
  }
}