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.
def f1():
	t=int(input())
	while(t!=0):
		t=t-1
		s=str(input())
		c1,c0=count(s)
		if(c1==0 or c0==0):
			print(s)
			#print()
			continue
		L=len(s)
		if(s[0]=="0"):
			ans=["0" if(i%2==0) else "1" for i in range(0,2*len(s))]
		else:
			ans=["1" if(i%2==0) else "0" for i in range(0,2*len(s))]
		ans="".join(ans)	

		#ans=""+str(s[0])
		#for i in range(1,len(s)):
		#	if(s[i]==s[i-1]):
		#		ans=ans+NT(s[i])+s[i]
		#	else:
		#		ans=ans+s[i]
		#		if(i==len(s)-1 and len(ans)%2!=0):
		#			ans=ans+NT(s[i])
		print(ans)
		#print()

def NT(strn):
	if(strn=="0"):
		return "1"
	else:
		return "0"
def count(strn):
	c1,c0=0,0
	for i in range(0,len(strn)):
		if(str(strn[i])==str("0")):
			c0+=1
		else:
			c1+=1
	return c1,c0
if __name__ == '__main__':
	f1()