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 long long N = (long long)(1e6) + 10;
const long long mod = 998244353;
const long long inf = 1e9;
const double eps = 1e-16;
const double pi = acos(-1);
inline long long rd() {
  long long p = 0;
  long long f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f *= -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    p = p * 10 + ch - '0';
    ch = getchar();
  }
  return p * f;
}
char s[N], t[N], ans[N];
signed main() {
  long long tcase = rd();
  while (tcase--) {
    scanf("%s", s + 1);
    long long len = strlen(s + 1);
    long long ansl = 2 * len;
    for (long long i = 1; i <= len; i++) ans[i] = ans[i + len] = s[i];
    for (long long i = 1; i <= len; i++)
      for (long long j = i; j <= len; j++) {
        long long l = j - i + 1;
        for (long long p = 1; p <= l; p++) t[p] = s[i + p - 1];
        long long ti = 2 * len / l;
        long long cir = l;
        for (long long p = 1; p <= cir * ti; p++) l++, t[l] = t[l - cir];
        long long k = 1;
        for (long long p = 1; p <= l; p++)
          if (t[p] == s[k] && k <= len) k++;
        if (k == len + 1 && cir < ansl) {
          ansl = cir;
          for (long long p = 1; p <= l; p++) ans[p] = t[p];
        }
      }
    long long ti = 2 * len / ansl;
    for (long long i = 1; i <= ansl * ti; i++) printf("%c", ans[i]);
    puts("");
  }
  return 0;
}
