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;
const long long N = (long long)5e5 + 2;
const long long MOD = (long long)1e9 + 7;
const long long MOD1 = (long long)998244353;
const long long inf = (long long)1e18 + 6;
template <class T>
bool uin(T &a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool uax(T &a, T b) {
  return a < b ? (a = b, true) : false;
}
long long add(long long x, long long y) {
  x += y;
  if (x >= MOD) return x - MOD;
  return x;
}
long long sub(long long x, long long y) {
  x -= y;
  if (x < 0) return x + MOD;
  return x;
}
long long mult(long long x, long long y) { return (x * y) % MOD; }
long long bin_pow(long long x, long long p) {
  if (p == 0) return 1;
  if (p & 1) return mult(x, bin_pow(x, p - 1));
  return bin_pow(mult(x, x), p / 2);
}
long long rev(long long x) { return bin_pow(x, MOD - 2); }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.precision(20);
  cout << fixed;
  long long t;
  cin >> t;
  while (t--) {
    string s;
    long long cnt0 = 0, cnt1 = 0;
    cin >> s;
    for (int i = 0; i < (int)(s.length()); ++i)
      if (s[i] == '0')
        cnt0++;
      else
        cnt1++;
    if (cnt1 == 0)
      for (int i = 0; i < (int)(cnt0); ++i) cout << 0;
    else if (cnt0 == 0)
      for (int i = 0; i < (int)(cnt1); ++i) cout << 1;
    else {
      for (int i = 0; i < (int)(s.length()); ++i) cout << 0 << 1;
    }
    cout << "\n";
  }
}