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 T;
  cin >> T;
  while (T--) {
    string a;
    cin >> a;
    long long int i, j, k;
    char b[2 * a.length()];
    bool f = 0;
    for (i = 1; i < 2 * a.length() && !f; i++) {
      for (j = 0; j <= a.length(); j++) b[j] = '0';
      for (j = 0; j < 2 * a.length(); j += i) b[j] = '1';
      k = 0;
      for (j = 0; j < 2 * a.length() && !f; j++) {
        if (b[j] == a[k]) {
          k++;
          if (k == a.length()) f = 1;
        }
      }
      if (!f) {
        for (j = 0; j < 2 * a.length(); j++) b[j] = '1';
        for (j = 0; j < 2 * a.length(); j += i) b[j] = '0';
        k = 0;
        for (j = 0; j < 2 * a.length() && !f; j++) {
          if (b[j] == a[k]) {
            k++;
            if (k == a.length()) f = 1;
          }
        }
      }
    }
    b[2 * a.length()] = '\0';
    cout << b << endl;
  }
  return 0;
}