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.
#include <bits/stdc++.h>
using namespace std;
int main() {
  int cases;
  cin >> cases;
  for (int i = 0; i < cases; i++) {
    bool isSame = true;
    string str;
    cin >> str;
    for (int j = 1; j < str.length(); j++) {
      if (str[j - 1] == str[j])
        continue;
      else
        isSame = false;
    }
    if (isSame)
      cout << str << endl;
    else {
      if (str.length() <= 2)
        cout << str << endl;
      else {
        string testStr = str;
        int one0Count = 0;
        int zero1Count = 0;
        while (str.find("10") != string::npos) {
          one0Count += 1;
          str[str.find("10")] = ' ';
          str[str.find("10") + 1] = ' ';
        }
        while (str.find("01") != string::npos) {
          zero1Count += 1;
          str[str.find("01")] = ' ';
          str[str.find("01") + 1] = ' ';
        }
        if (zero1Count > one0Count) {
          int num2 = str.length() - zero1Count;
          for (int k = 0; k < num2 - 1; k++) cout << "01";
          cout << "01" << endl;
        } else if (zero1Count <= one0Count) {
          int num2 = str.length() - one0Count;
          for (int l = 0; l < num2 - 1; l++) cout << "10";
          cout << "10" << endl;
        }
      }
    }
  }
}
