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;
long long div_up(long long x, long long y) {
  if (x == 0) return 0;
  return (x - 1) / y + 1;
}
bool is_prime(long long a) {
  for (long long i = 2; i <= sqrt(a); i++)
    if (a % i == 0) return false;
  return true;
}
bool borders(int x, int y, int n, int m) {
  return x >= 0 && y >= 0 && x < n && y < m;
}
void solve() {
  string s;
  cin >> s;
  if (s.find('0') == string::npos || s.find('1') == string::npos)
    cout << s << '\n';
  else {
    for (int i = 0; i < s.length() - 1; i++) {
      if (s[i] == s[i + 1]) {
        if (s[i] == '0')
          s.insert(i + 1, "1");
        else
          s.insert(i + 1, "0");
        i++;
      }
    }
    cout << s << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie();
  cout.tie();
  int T;
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}