Let's say string s has period k if s_i = s_{i + k} for all i from 1 to |s| - k (|s| means length of string s) and k is the minimum positive integer with this property.

Some examples of a period: for s="0101" the period is k=2, for s="0000" the period is k=1, for s="010" the period is k=2, for s="0011" the period is k=4.

You are given string t consisting only of 0's and 1's and you need to find such string s that:

  1. String s consists only of 0's and 1's; 
  2. The length of s doesn't exceed 2 ⋅ |t|; 
  3. String t is a subsequence of string s; 
  4. String s has smallest possible period among all strings that meet conditions 1—3. 



Let us recall that t is a subsequence of s if t can be derived from s by deleting zero or more elements (any) without changing the order of the remaining elements. For example, t="011" is a subsequence of s="10101".

Input

The first line contains single integer T (1 ≤ T ≤ 100) — the number of test cases.

Next T lines contain test cases — one per line. Each line contains string t (1 ≤ |t| ≤ 100) consisting only of 0's and 1's.

Output

Print one string for each test case — string s you needed to find. If there are multiple solutions print any one of them.

Example

Input


4
00
01
111
110


Output


00
01
11111
1010

Note

In the first and second test cases, s = t since it's already one of the optimal solutions. Answers have periods equal to 1 and 2, respectively.

In the third test case, there are shorter optimal solutions, but it's okay since we don't need to minimize the string s. String s has period equal to 1.
#include <bits/stdc++.h>
using namespace std;
template <class T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  for (T i : v) os << i << ' ';
  return os;
}
template <class T>
ostream &operator<<(ostream &os, const set<T> &v) {
  for (T i : v) os << i << ' ';
  return os;
}
template <class T, class S>
ostream &operator<<(ostream &os, const pair<T, S> &v) {
  os << v.first << ' ' << v.second;
  return os;
}
template <class T, class S>
ostream &operator<<(ostream &os, const unordered_map<T, S> &v) {
  for (auto i : v) os << '(' << i.first << "=>" << i.second << ')' << ' ';
  return os;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int T;
  cin >> T;
  for (int ti = 1; ti <= T; ti++) {
    string t;
    cin >> t;
    bool f = true;
    char a = t.at(0);
    for (int i = int(1); i <= int(t.size() - 1); i++)
      if (t.at(i) != a) {
        f = false;
        break;
      }
    if (f)
      cout << t << "\n";
    else {
      int l = 2 * (t.size());
      if (a == 0)
        for (int i = int(0); i <= int(l - 1); i++) {
          if (i % 2 == 0)
            cout << 0;
          else
            cout << 1;
        }
      else
        for (int i = int(0); i <= int(l - 1); i++) {
          if (i % 2 == 0)
            cout << 1;
          else
            cout << 0;
        }
      cout << "\n";
    }
  }
  return 0;
}
