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

def stringslicing(string,index):
	if index>=0 and index<len(string):
		return string[index]
	else:
		return "not defined"

t=int(input())
for _ in range(t):
	string=input().strip()
	s=int(input())
	answer={}
	answer=answer.fromkeys(range(0,len(string)),"1")
	
	
	for i in range(len(string)):
		if string[i]=="0":
			a=i-s
			b=i+s
			if a>=0 and a<len(string):
				answer[a]="0"
			if b>=0 and b<len(string):
				answer[b]="0"
	
	ans="".join(answer.values())

	wrong=0
	for k in range(len(string)):
		if string[k]=="1":
			if string[k]==stringslicing(ans,k+s) or string[k]==stringslicing(ans, k-s):
				continue
			else:
				print(-1)
				wrong=1
				break

	if wrong==0:
		print(ans)

	