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;
mt19937_64 mt(chrono::steady_clock::now().time_since_epoch().count());
inline long long S(int n, int i = 1, int d = 1) {
  return n & 1 ? (2 * i + d * (n - 1)) / 2 * n : n / 2 * (2 * i + d * (n - 1));
}
inline string ctos(char c) { return string(1, c); }
inline int popcnt(long long x) {
  int ans = 0;
  for (int i = 0; i < 64; i++)
    if ((x >> i) & 1) ans++;
  return ans;
}
const int SEED = 1e9 + 7 + 7e7;
const int SIZE = 2e5 + 17;
const int M = 1000001917;
const int LOG = 18;
const int INF = 1e9 + 17;
long long gcd(long long a, long long b) { return a % b ? gcd(b, a % b) : b; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int _;
  cin >> _;
  while (_--) {
    string t;
    cin >> t;
    set<char> s;
    for (int i = 0; i < t.size(); i++) s.insert(t[i]);
    for (int i = 0; i < 2 * t.size(); i++) {
      auto it = s.begin();
      advance(it, i % s.size());
      cout << *it;
    }
    cout << endl;
  }
}
