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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long int dx[] = {-1, 0, 1, 0};
long long int dy[] = {0, -1, 0, 1};
void c_p_c() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
bool is_pallindrome(string str, long long int left, long long int right) {
  if (left > right) return true;
  return (str[left] == str[right] && is_pallindrome(str, left + 1, right - 1));
}
long long int convertToInt(string str) {
  long long int sum = 0;
  long long int power = 1;
  for (long long int i = str.length() - 1; i >= 0; i--) {
    long long int val = str[i] - '0';
    long long int prod =
        ((val % 1000000007) * (power % 1000000007)) % 1000000007;
    sum = ((sum % 1000000007) + prod % 1000000007) % 1000000007;
    power *= 10;
  }
  return sum;
}
bool compare(pair<long long int, long long int> a,
             pair<long long int, long long int> b) {
  return a.first > b.first;
}
int32_t main() {
  c_p_c();
  long long int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    long long int i;
    long long int period = 1;
    for (i = 1; i < s.length(); i++) {
      if (s[i] != s[i - 1]) {
        period = 2;
        break;
      }
    }
    if (period == 1)
      cout << s;
    else {
      if (s[0] == '1') {
        for (long long int i = 1; i <= s.length(); i++) cout << "10";
      } else {
        for (long long int j = 1; j <= s.length(); j++) cout << "01";
      }
    }
    cout << "\n";
  }
  return 0;
}