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;
string result(string s, int cnt1, int cnt0) {
  if (cnt0 == 0) {
    return s;
  }
  string res = "";
  for (int i = 0; i <= s.size() - 1; i++) {
    if (s[i] == '1') {
      if (i != s.size() - 1 && s[i + 1] == '1') {
        res += "10";
      } else
        res += '1';
    } else {
      if (i != s.size() - 1 && s[i + 1] == '0') {
        res += "01";
      } else
        res += '0';
    }
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  for (int i = 1; i <= t; i++) {
    string str;
    cin >> str;
    int cnt0 = 0;
    int cnt1 = 0;
    int rev = 0;
    for (int j = 0; j <= str.size() - 1; j++) {
      if (str[j] == '0') {
        cnt0++;
      } else
        cnt1++;
    }
    if (cnt0 > cnt1) {
      rev = 1;
      swap(cnt1, cnt0);
      for (int j = 0; j <= str.size() - 1; j++) {
        if (str[j] == '0') {
          str[j] = '1';
        } else
          str[j] = '0';
      }
    }
    string res = result(str, cnt1, cnt0);
    if (rev) {
      for (int i = 0; i <= res.size() - 1; i++) {
        res[i] = (res[i] == '0') ? '1' : '0';
      }
    }
    cout << res << "\n";
  }
  return 0;
}