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;
class algoritm {
  friend istream &operator>>(istream &, algoritm &);
  friend ostream &operator<<(ostream &, const algoritm &);

 public:
  void rezolvare() {
    cin.get();
    for (int i = 1; i <= n; i++) {
      string s;
      getline(cin, s);
      int nr1, nr2;
      nr1 = 0;
      nr2 = 0;
      for (int j = 0; j < s.size(); j++) {
        if (s[j] == '1') {
          nr1++;
        } else if (s[j] == '0') {
          nr2++;
        }
      }
      if (nr1 == s.size()) {
        cout << s << endl;
        continue;
      } else if (nr2 == s.size()) {
        cout << s << endl;
        continue;
      }
      for (int j = 0; j < s.size() - 1; j++) {
        if ((s[j] == '1' && s[j + 1] == '0') ||
            (s[j] == '1' && s[j + 1] == '0')) {
          continue;
        } else if (s[j] == '1' && s[j + 1] == '1') {
          s.insert(j + 1, "0");
        } else if (s[j] == '0' && s[j + 1] == '0') {
          s.insert(j + 1, "1");
        }
      }
      cout << s << endl;
    }
  }

 protected:
  int n;
};
ostream &operator<<(ostream &output, const algoritm &alg) { return output; }
istream &operator>>(istream &input, algoritm &alg) {
  input >> alg.n;
  return input;
}
int main() {
  algoritm alg;
  cin >> alg;
  alg.rezolvare();
}
