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;
inline int qr() {
  int f = 0, fu = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') fu = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    f = (f << 3) + (f << 1) + c - 48;
    c = getchar();
  }
  return f * fu;
}
const int N = 105;
char t[N], s[N << 1];
int T;
int main() {
  T = qr();
  while (T--) {
    scanf("%s", t + 1);
    int n = strlen(t + 1);
    if (count(t + 1, t + 1 + n, '0') == 0 || count(t + 1, t + 1 + n, '1') == 0)
      printf("%s\n", t + 1);
    else {
      int now = 1;
      for (register int i = 1; i <= n - 1; ++i) {
        if (t[i] == t[i + 1]) {
          if (t[i] == '1')
            s[now] = '1', s[now + 1] = '0';
          else
            s[now] = '0', s[now + 1] = '1';
          now += 2;
        } else
          s[now++] = t[i];
      }
      if (t[n] == t[1]) {
        if (t[n] == '1')
          s[now] = '1', s[now + 1] = '0';
        else
          s[now] = '0', s[now + 1] = '1';
        now += 2;
      } else
        s[now++] = t[n];
      s[now] = '\0';
      printf("%s\n", s + 1);
    }
  }
  return 0;
}
