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;
const int INF = (1 << 30);
const long long INFLL = (1ll << 60);
const long double PI = 3.141592653589793238L;
template <typename T>
void pvv(T a, T b) {
  for (T i = a; i != b; ++i) cerr << *i << " ";
  cerr << endl;
}
template <typename T>
void pv(vector<T> v) {
  for (T x : v) cerr << x << " ";
  cerr << endl;
}
template <typename T>
void chmin(T& a, T b) {
  if (a > b) a = b;
}
template <typename T>
void chmax(T& a, T b) {
  if (a < b) a = b;
}
int in() {
  int x;
  scanf("%d", &x);
  return x;
}
double fin() {
  double x;
  scanf("%lf", &x);
  return x;
}
long long lin() {
  long long x;
  scanf("%lld", &x);
  return x;
}
void doStuff() {
  string t;
  cin >> t;
  bool change = false;
  char prev = t[0];
  string s(1, prev);
  for (int i = 0; i < (int)(t.size() - 1); i++) {
    if (t[i + 1] == prev)
      s += (char)('0' + '1' - prev);
    else
      change = true, prev = t[i + 1];
    s += prev;
  }
  if (change)
    cout << s << endl;
  else
    cout << t << endl;
}
int main() {
  int T = in();
  for (int t = 0; t < (int)(T); t++) doStuff();
  return 0;
}
