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;
long long int bigMod(long long int x, long long int y,
                     long long int MOD = 10000007) {
  if (y == 0) return 1 % MOD;
  long long int haha = bigMod(x, y / 2, MOD);
  haha = (haha * haha) % MOD;
  if (y & 1) return (haha * x) % MOD;
  return haha;
}
long long int modInverse(long long int x, long long int MOD) {
  return bigMod(x, MOD - 2, MOD);
}
long long int sieveMama() {
  bool isPrime[1000000];
  memset(isPrime, true, sizeof(isPrime));
  isPrime[1] = false;
  long long int n;
  cin >> n;
  vector<long long int> v;
  for (long long int i = 2; i <= n; i += 2) isPrime[i] = false;
  for (long long int i = 3; i * i <= n; i += 2) {
    if (isPrime[i]) {
      for (long long int m = i * i; m <= n; m += i) {
        isPrime[m] = false;
      }
    }
  }
}
vector<long long int> primeFactorize(long long int n) {
  vector<long long int> primeFactors;
  for (long long int i = 2; i * i <= n; i++) {
    while (n % i == 0) {
      n = n / i;
      primeFactors.push_back(i);
    }
  }
  if (n > 1) primeFactors.push_back(n);
  return primeFactors;
}
int binarySearch(int arr[], int lo, int hi, int x) {
  if (hi >= lo) {
    int mid = lo + (hi - lo) / 2;
    if (arr[mid] == x) return mid;
    if (arr[mid] > x) return binarySearch(arr, lo, mid - 1, x);
    return binarySearch(arr, mid + 1, hi, x);
  }
  return -1;
}
int main() {
  long long int t;
  cin >> t;
  while (t--) {
    long long int one = 0, zero = 0;
    string s, s2 = "";
    cin >> s;
    for (long long int i = 0; i < s.size(); i++) {
      if (s[i] == '1')
        one++;
      else
        zero++;
    }
    if (one == s.size() || zero == s.size())
      cout << s << endl;
    else {
      for (long long int i = 1; i <= s.size() * 2; i++) {
        if (i % 2 == 1)
          s2 += "1";
        else
          s2 += "0";
      }
      cout << s2 << endl;
    }
  }
}