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;
long long int MOD = 998244353;
long long int pwr(long long int x, long long int y) {
  long long int res = 1;
  x = x % MOD;
  while (y > 0) {
    if (y & 1) res = (res * x) % MOD;
    y = y >> 1;
    x = (x * x) % MOD;
  }
  return res;
}
inline long long int addmod(long long int a, long long int b) {
  return ((a + b) % MOD);
}
inline long long int mulmod(long long int a, long long int b) {
  return ((a * b) % MOD);
}
class BIT {
  int *bit;
  int N;

 public:
  void update(int x, int val) {
    int ind = x;
    while (ind <= N) {
      bit[ind] += val;
      ind += (ind & (-ind));
    }
  }
  BIT(int ar[], int n) {
    bit = new int[n + 1];
    N = n + 1;
    for (int i = 1; i < N; i++) bit[i] = 0;
    for (int i = 1; i < N; i++) update(i, ar[i - 1]);
  }
  int getSum(int x) {
    if (x < 0) return 0;
    int ind = x + 1;
    int sum = 0;
    while (ind > 0) {
      sum += bit[ind];
      ind = (ind & (ind - 1));
    }
    return sum;
  }
  int getValue(int x) { return getSum(x) - getSum(x - 1); }
  void changeElem(int x, int val) { update(x + 1, val - getValue(x)); }
};
int main() {
  int T;
  cin >> T;
  while (T--) {
    string s, t;
    cin >> t;
    bool allsame = true;
    for (int i = 0; i < t.size(); i++) {
      s += "01";
      if (i > 0) {
        if (t[i] != t[i - 1]) allsame = false;
      }
    }
    if (allsame)
      cout << t << endl;
    else
      cout << s << endl;
  }
}
