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;
void canhazfast() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
}
template <typename T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <typename T>
T extgcd(T a, T b, T &x, T &y) {
  T x0 = 1, y0 = 0, x1 = 0, y1 = 1;
  while (b) {
    T q = a / b;
    a %= b;
    swap(a, b);
    x0 -= q * x1;
    swap(x0, x1);
    y0 -= q * y1;
    swap(y0, y1);
  }
  x = x0;
  y = y0;
  return a;
}
int ctz(unsigned x) { return __builtin_ctz(x); }
int ctzll(unsigned long long x) { return __builtin_ctzll(x); }
int clz(unsigned x) { return __builtin_clz(x); }
int clzll(unsigned long long x) { return __builtin_clzll(x); }
int popcnt(unsigned x) { return __builtin_popcount(x); }
int popcntll(unsigned long long x) { return __builtin_popcountll(x); }
int bsr(unsigned x) { return 31 ^ clz(x); }
int bsrll(unsigned long long x) { return 63 ^ clzll(x); }
char s[2 * 128], t[128];
void solve() {
  int n;
  cin >> t;
  n = strlen(t);
  for (int i = 0; i < 2 * n; ++i) s[i] = i % 2 + '0';
  s[2 * n] = 0;
  for (int i = 1; i < n; ++i) {
    if (t[i] != t[0]) return cout << s << '\n', void();
  }
  cout << t << '\n';
}
int main() {
  canhazfast();
  int t;
  cin >> t;
  for (; t; --t) solve();
  return 0;
}
