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() {
  long long tc;
  tc = 1;
  cin >> tc;
  while (tc--) {
    string s;
    cin >> s;
    string x = s;
    string y = "";
    long long f = 1;
    char a = x[0];
    for (long long i = 1; i < x.length(); i++) {
      if (x[i] != a) {
        f = 0;
        break;
      }
    }
    if (f == 1)
      cout << x << "\n";
    else {
      a = x[0];
      y = y + x[0];
      for (long long i = 1; i < x.length(); i++) {
        if (a == x[i]) {
          if (a == '0') {
            y += '1';
            y += x[i];
          } else {
            y += '0';
            y += x[i];
          }
        } else {
          y += x[i];
          if (x[i] == '0')
            a = '0';
          else
            a = '1';
        }
      }
      cout << y << "\n";
    }
  }
  return 0;
}