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.
#8:47
MOD=10**9+7
INT_MAX=10**20+7
def INPUT():return list(int(i) for i in input().split())
def LIST_1D_ARRAY(n):return [0 for _ in range(n)]
def LIST_2D_ARRAY(m,n):return [[0 for _ in range(n)]for _ in range(m)]
def power(a,x):
    res=1
    while(x>0):
        if x&1:
            res=(res*a)%MOD
        a=(a*a)%MOD
        x//=2
    return res%MOD
def NCR_MOD(n,r):
    num=1
    for i in range(n,n-r,-1):num=(num*i)%MOD
    for j in range(1,r+1):num=(num*power(j,MOD-2))%MOD
    return num

#################################################################################
def period(s):
    for p in range(1,len(s)+1):
        bool=True
        if len(s)%p==0:
            for i in range(len(s)-p):
                if s[i]!=s[i+p]:
                    bool=False
            if bool:
                return p

def calc(s):
    ans=""
    p=period(s)
    if p==1 or p==2:
        return s
    # ans+=s[0]
    # for i in range(1,len(s)):
    #     if int(s[i])+int(ans[-1])==1:
    #         ans+=str(s[i])
    #     else:
    #         ans+=str(1-f)+str(s[i])
    ans="01"*len(s)
    return ans

for _ in range(int(input())):
    s=input()
    print(calc(s))
