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>
using namespace std;
const long long int logx = 30;
const long long int N = 1e5 + 5;
const long long int M = 3e5 + 5;
const long long int mod = 1e9 + 7;
const long long int INF = 1e18 + 5;
const double PI = 3.14159265358979323846;
inline long long int mul(long long int a, long long int b) {
  return (a * 1ll * b) % mod;
}
inline long long int sub(long long int a, long long int b) {
  long long int c = a - b;
  if (c < 0) c += mod;
  return c;
}
inline long long int add(long long int a, long long int b) {
  long long int c = a + b;
  if (c > mod) c -= mod;
  return c;
}
template <typename T>
T power(T x, T y, long long int m) {
  T ans = 1;
  while (y > 0) {
    if (y & 1LL) ans = (ans * x) % m;
    y >>= 1ll;
    x = (x * x) % m;
  }
  return ans % m;
}
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
long long int n, m;
string second, t;
long long int dp[205][205][410];
long long int F(long long int i, long long int j, long long int open) {
  long long int res = INF;
  if (open < 0) {
    return INF;
  }
  if (open >= 410) {
    return INF;
  }
  if (open == 0) {
    if (i == n && j == m) {
      return 0;
    }
  }
  if (dp[i][j][open] != -1) {
    return dp[i][j][open];
  }
  if (i == n && j == m) {
    return open;
  }
  if (i == n && j != m) {
    res = min(res, 1 + F(i, j + (t[j] == '('), open + 1));
    res = min(res, 1 + F(i, j + (t[j] == ')'), open - 1));
  }
  if (i != n && j == m) {
    res = min(res, 1 + F(i + (second[i] == '('), j, open + 1));
    res = min(res, 1 + F(i + (second[i] == ')'), j, open - 1));
  }
  if (i != n && j != m) {
    res = min(res, 1 + F(i + (second[i] == '('), j + (t[j] == '('), open + 1));
    res = min(res, 1 + F(i + (second[i] == ')'), j + (t[j] == ')'), open - 1));
  }
  return dp[i][j][open] = res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  memset(dp, -1, sizeof(dp));
  cin >> second >> t;
  n = second.length();
  m = t.length();
  string out = "\0";
  long long int ans = F(0, 0, 0), prevans = 0, prei = 0, prej = 0, preopen = 0;
  for (long long int i = 0; i < ans; i++) {
    long long int tempans =
        1 + prevans +
        F(prei + (second[prei] == '('), prej + (t[prej] == '('), preopen + 1);
    if (tempans == ans) {
      out += "(";
      prevans++;
      preopen++;
      prei += (second[prei] == '(');
      prej += (t[prej] == '(');
      continue;
    }
    tempans =
        1 + prevans +
        F(prei + (second[prei] == ')'), prej + (t[prej] == ')'), preopen - 1);
    if (tempans == ans) {
      out += ")";
      prevans++;
      preopen--;
      prei += (second[prei] == ')');
      prej += (t[prej] == ')');
      continue;
    }
  }
  cout << out << endl;
}