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 int MAXN = 1000005;
const int SQRTN = 1003;
const int LOGN = 22;
const double PI = acos(-1);
const int INF = 1e9;
const int MOD = 1000000007;
const int FMOD = 998244353;
const double eps = 1e-9;
template <typename T>
T gcd(T a, T b) {
  return (b ? __gcd(a, b) : a);
}
template <typename T>
T lcm(T a, T b) {
  return (a * (b / gcd(a, b)));
}
template <typename T>
T add(T a, T b, T c = MOD) {
  T res = a + b;
  return (res >= c ? res - c : res);
}
template <typename T>
T mod_neg(T a, T b, T c = MOD) {
  T res;
  if (abs(a - b) < c)
    res = a - b;
  else
    res = (a - b) % c;
  return (res < 0 ? res + c : res);
}
template <typename T>
T mul(T a, T b, T c = MOD) {
  long long res = (long long)a * b;
  return (res >= c ? res % c : res);
}
inline long long mulmod(long long a, long long b, long long m = MOD) {
  long long q = (long long)(((long double)a * (long double)b) / (long double)m);
  long long r = a * b - q * m;
  if (r > m) r %= m;
  if (r < 0) r += m;
  return r;
}
template <typename T>
T expo(T e, T n) {
  T x = 1, p = e;
  while (n) {
    if (n & 1) x = x * p;
    p = p * p;
    n >>= 1;
  }
  return x;
}
template <typename T>
T power(T e, T n, T m = MOD) {
  T x = 1, p = e;
  while (n) {
    if (n & 1) x = mul(x, p, m);
    p = mul(p, p, m);
    n >>= 1;
  }
  return x;
}
template <typename T>
T extended_euclid(T a, T b, T &x, T &y) {
  T xx = 0, yy = 1;
  y = 0;
  x = 1;
  while (b) {
    T q = a / b, t = b;
    b = a % b;
    a = t;
    t = xx;
    xx = x - q * xx;
    x = t;
    t = yy;
    yy = y - q * yy;
    y = t;
  }
  return a;
}
template <typename T>
T mod_inverse(T a, T n = MOD) {
  T x, y, z = 0;
  T d = extended_euclid(a, n, x, y);
  return (d > 1 ? -1 : mod_neg(x, z, n));
}
void it_2107() {
  long long k = 0, n = 0, x = 0, m = 0, curr = 0, num = 0, n1 = 0, n2 = 0,
            n3 = 0, n4 = 0, i = 0, j = 0;
  long long root = 0, sum = 0, diff = 0, q = 0, choice = 0, d = 0, len = 0,
            beg = 0, end = 0, pos = 0, cnt = 0;
  long long lo = 0, hi = 0, mid = 0, ans = 0;
  bool flag = false;
  string s1, s2, s3, str, t;
  char ch, ch1, ch2, ch3, *ptr;
  double dub = 0;
  cin >> t;
  len = t.length();
  long long cntt[] = {0, 0};
  for (int(i) = 0; (i) < (len); (i)++) {
    cntt[t[i] - '0']++;
  }
  if (cntt[1] == 0 || cntt[0] == 0) {
    cout << t << "\n";
  } else {
    for (int(i) = 0; (i) < (len); (i)++) {
      cout << "10";
    }
    cout << "\n";
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t = 1;
  cin >> t;
  while (t--) {
    it_2107();
  }
  return 0;
}
