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.
# for t in range(int(input())):
#     s = input()
#     i, j = 0, 0
#     cnt = 0
#     ans = float('inf')
#     dic = {}
#     while j < len(s):
#         if len(dic) < 3:
#             dic[s[j]] = dic.get(s[j], 0) + 1
#         # print(j)
#         # print(dic)
#         while len(dic) == 3:
#             ans = min(ans, j - i + 1)
#             dic[s[i]] -= 1
#             if dic[s[i]] == 0:
#                 del dic[s[i]]
#             i += 1
#
#         j += 1
#     print((0, ans)[ans < float('inf')])


# for t in range(int(input())):
#     n = int(input())
#     s = list(map(int, input().split()))
#     dp = [1] * n
#     for i in range(n):
#         k = 2
#         while (i + 1) * k <= n:
#             j = (i + 1) * k
#             if s[i] < s[j - 1]:
#                 dp[j - 1] = max(dp[j - 1], dp[i] + 1)
#             k += 1
#     print(max(dp))


for T in range(int(input())):
    t = input()
    z, o = 0, 0
    for ch in t:
        z = z + 1 if ch == '0' else z
        o = o + 1 if ch == '1' else o
    if z > 0 and o > 0:
        print('01' * len(t))
    elif o > 0 and not z:
        print('1' * len(t))
    else:
        print('0' * len(t))
