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;
signed main() {
  long long t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    long long n = s.length();
    long long max_diff1 = 0;
    long long max_diff2 = 0;
    long long prev_0 = -1, next_0 = -1;
    long long prev_1 = -1, next_1 = -1;
    for (long long i = 0; i < n; i++) {
      if (s[i] == '0') {
        if (prev_0 == -1) {
          prev_0 = i;
        } else {
          if (next_0 == -1) {
            next_0 = i;
            max_diff1 = max(max_diff1, next_0 - prev_0);
          } else {
            prev_0 = next_0;
            next_0 = i;
            max_diff1 = max(max_diff1, next_0 - prev_0);
          }
        }
      } else {
        if (prev_1 == -1) {
          prev_1 = i;
          max_diff2 = max(max_diff2, next_1 - prev_1);
        } else {
          if (next_1 == -1) {
            next_1 = i;
          } else {
            prev_1 = next_1;
            next_1 = i;
            max_diff2 = max(max_diff2, next_1 - prev_1);
          }
        }
      }
    }
    if (prev_0 == -1 || prev_1 == -1)
      cout << s << endl;
    else if (next_0 == -1 && next_1 == -1)
      cout << s << endl;
    else {
      vector<char> ans;
      for (long long i = 0; i < n; i++) {
        if (s[i] == '0') {
          ans.push_back(s[i]);
          if (i + 1 < n && s[i + 1] == '0') ans.push_back('1');
        } else {
          ans.push_back(s[i]);
          if (i + 1 < n && s[i + 1] == '1') ans.push_back('0');
        }
      }
      for (long long i = 0; i < ans.size(); i++) cout << ans[i];
      cout << endl;
    }
  }
}