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;
inline long long isqrt(long long k) {
  long long r = sqrt(k) + 1;
  while (r * r > k) r--;
  return r;
}
inline long long icbrt(long long k) {
  long long r = cbrt(k) + 1;
  while (r * r * r > k) r--;
  return r;
}
inline long long mnz(long long& a, long long b) { return a = (a > b ? b : a); }
inline long long mxz(long long& a, long long b) { return a = (a < b ? b : a); }
inline string toString(long long n) {
  stringstream ss;
  ss << n;
  return ss.str();
}
double const eps = 1e-6;
long long const Base = 1e9 + 7, oo = 1e17, MAXN = 1e6;
long long A[MAXN + 5];
long long Solves() {
  long long n, m, k, cnt = 0, ans = 0, x, y, q, c, sum = 0, v, t;
  string s;
  cin >> s;
  for (long long i = 0; i + 1 <= s.length(); i++)
    if (i and s[i] != s[i - 1]) ans = 1;
  if (!ans) return cout << s << endl, 0;
  for (long long i = 0; i + 1 <= s.length(); i++) {
    if (s[i] == ans + '0')
      cout << (char)(ans + '0');
    else {
      cout << (char)(ans + '0');
      cout << s[i];
      ans = s[i] - '0';
    }
    ans ^= 1;
  }
  cout << "" << endl;
  return 0;
}
int main() {
  if (fopen("locin.txt", "r")) freopen("locin.txt", "r", stdin);
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long test = 1;
  cin >> test;
  for (long long T = 1; T <= test; T++) {
    Solves();
  }
}