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;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int INF = 1000000000;
const int mod = 1e9 + 7;
const int maxn = 2e5 + 5;
string s;
int check() {
  int _0 = 0, _1 = 0;
  for (int i = 0; i < s.size(); ++i) {
    if (s[i] == '0')
      _0++;
    else
      _1++;
  }
  return (!_0 || !_1);
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    cin >> s;
    if (check())
      cout << s << "\n";
    else {
      printf("%c", s[0]);
      for (int i = 1; i < s.size(); ++i) {
        if (s[i] == s[i - 1]) printf("%c", s[i] == '1' ? '0' : '1');
        printf("%c", s[i]);
      }
      printf("\n");
    }
  }
  return 0;
}