Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:

  * if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} =  1, then s_i =  1); 
  * if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} =  1, then s_i =  1); 
  * if both of the aforementioned conditions are false, then s_i is 0. 



You are given the integer x and the resulting string s. Reconstruct the original string w.

Input

The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.

Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).

The total length of all strings s in the input does not exceed 10^5.

Output

For each test case, print the answer on a separate line as follows:

  * if no string w can produce the string s at the end of the process, print -1; 
  * otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them. 

Example

Input


3
101110
2
01
1
110
1


Output


111011
10
-1
t=int(input())
for _ in range(t):
	s=input()
	x=int(input())
	n=len(s)
	ans=''
	flag=1
	for i in range(n):
		if(i<x and i<(n-x)):
			if(s[i+x]=='1'):
				ans+='1'
			else:
				ans+='0'
		elif(i>=(n-x) and i>=x):
			if(s[i-x]=='1'):
				ans+='1'
			else:
				ans+='0'
		elif(i-x<0 and i+x>=n):
			ans+='0'
		else:
			if(s[i-x]=='0' and s[i+x]=='0'):
				ans+='0'
			elif(s[i-x]=='1' and s[i+x]=='1'):
				ans+='1'
			else:
				ans+='0'
	for i in range(n):
		if(s[i]=='0'):
			if(i<x and i<(n-x)):
				if(ans[i+x]!='0'):
					flag=0
					break
			elif(i>=(n-x) and i>=x):
				if(ans[i-x]!='0'):
					flag=0
					break
			elif(i-x<0 and i+x>=0):
				pass
			else:
				if(ans[i+x]=='1' or ans[i-x]=='1'):
					flag=0
					break
		else:
			if(i<x and i<(n-x)):
				if(ans[i+x]!='1'):
					flag=0
					break
			elif(i>=(n-x) and i>=x):
				if(ans[i-x]!='1'):
					flag=0
					break
			elif(i-x<0 and i+x>=0):
				flag=0
				break
			else:
				if(ans[i+x]=='0' and ans[i-x]=='0'):
					flag=0
					break
	if(flag):
		print(ans)
	else:
		print(-1)