Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main(int argc, char const *argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    int l = s.length();
    int t = 1;
    for (int i = 0; i < l - 1; i++) {
      if (s[i] != s[i + 1]) t = 2;
    }
    for (int i = 0; i < l - 2; i++) {
      if (l != 2 && s[i] != s[i + 2]) t = 3;
    }
    int la = 0;
    char ans[200];
    if (t == 3) {
      for (int i = 0; i < l - 1; ++i) {
        ans[la++] = s[i];
        if (s[i + 1] == s[i]) {
          if (s[i] == '0')
            ans[la++] = '1';
          else
            ans[la++] = '0';
        }
      }
      ans[la++] = s[l - 1];
      for (int i = 0; i < la; ++i) cout << ans[i];
      cout << endl;
    } else
      cout << s << endl;
  }
  return 0;
}