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 mod = 998244353;
int power(int a, int b) {
  long long c = 1;
  for (int i = 0; i < b; i++) c = c * a;
  return (c);
}
void reverseStr(string& string) {
  int n = string.length();
  for (int i = 0; i < n / 2; i++) swap(string[i], string[n - i - 1]);
}
int palindrom(string a, int m) {
  long long k = 1;
  for (long long i = 0; i < m / 2; i++) {
    cout << a[i] << " " << a[m - 1 - i];
    if (a[i] == a[m - 1 - i]) {
    } else {
      k = 0;
      break;
    }
  }
  if (k != 0)
    return (1);
  else
    return (0);
}
int main() {
  long long t;
  cin >> t;
  for (long long h = 0; h < t; h++) {
    string a;
    cin >> a;
    long long n = a.length();
    long long p = 0, f = 0;
    for (long long i = 0; i < n; i++) {
      if (a[i] == '0')
        p++;
      else
        f++;
    }
    if (p == n || f == n)
      cout << a << "\n";
    else {
      if (a[0] == '0') {
        for (long long i = 0; i < n; i++) cout << "01";
      } else {
        for (long long i = 0; i < n; i++) cout << "10";
      }
      cout << "\n";
    }
  }
}