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;
void printline(vector<int> v) {
  for (int i = 0; i < v.size(); i++) {
    if (i) cout << " ";
    cout << v[i];
  }
  cout << endl;
}
void printcol(vector<int> v) {
  for (auto x : v) cout << x << endl;
}
string s;
void build(int idx, int k, string base, string &ans, bool rev) {
  if (base.size() > (2 * s.size())) return;
  if (!ans.empty()) return;
  if (idx == (base.size() - k)) {
    if (rev) reverse((base).begin(), (base).end());
    ans = base;
    return;
  }
  if (base[idx] == base[idx + k]) {
    build(idx + 1, k, base, ans, rev);
  } else {
    string aux = base.substr(0, idx + k) + base[idx] +
                 base.substr(idx + k, base.size() - (idx + k));
    build(idx + 1, k, aux, ans, rev);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    cin >> s;
    map<char, int> freq;
    for (auto x : s) freq[x]++;
    string menor;
    if (s.size() <= 2 or freq.size() == 1) {
      menor = s;
    } else {
      char curr = s[0];
      menor += curr;
      curr = (curr == '0' ? '1' : '0');
      for (int i = 1; i < s.size(); i++) {
        if (curr != s[i]) i--;
        menor += curr;
        curr = (curr == '0' ? '1' : '0');
      }
    }
    cout << menor << endl;
  }
  return 0;
}