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
import sys
sys.setrecursionlimit(10000)
# default is 1000 in python

# increase stack size as well (for hackerrank)
# import resource
# resource.setrlimit(resource.RLIMIT_STACK, (resource.RLIM_INFINITY, resource.RLIM_INFINITY))


t = int(input())
# t = 1

for _ in range(t):
	s = input()
	x = int(input())
	n = len(s)

	sed = False
	w = ["0"]*n
	fixed = [0]*n

	for i in range(x):
		if i+x < n:
			w[i+x] = s[i]
			fixed[i+x] = 1
		elif s[i] == '1':
			sed = True
			break
	if sed is True:
		print("-1")
		continue
	sed = False
	# print(s)
	for i in range(x, n):
		# print(i)
		if s[i] == '1':
			if fixed[i-x] == 0 or w[i-x] == '1':
				w[i-x] = '1'
				fixed[i-x] = 1
			elif i+x < n and (fixed[i+x] == 0 or w[i+x] == '1'):
				w[i+x] = '1'
				fixed[i+x] = 1
			else:
				sed = True
				break
		else:
			if w[i-x] == '1' or (i+x < n and w[i+x] == '1'):
				sed = True
				break
			else:
				fixed[i-x] = 1
				if i+x < n:
					fixed[i+x] = 1

	if sed is True:
		print("-1")
	else:
		print(''.join(w))





# try:
	# raise Exception
# except:
	# print("-1")


# from itertools import combinations 
# all_combs = list(combinations(range(N), r))



# from collections import OrderedDict 
# mydict = OrderedDict() 


# thenos.sort(key=lambda x: x[2], reverse=True)


# int(math.log(max(numbers)+1,2))


# 2**3 (power)


# a,t = (list(x) for x in zip(*sorted(zip(a, t))))


# to copy lists use .copy()


# pow(p, si, 1000000007) for modular exponentiation


# my_dict.pop('key', None)
# This will return my_dict[key] if key exists in the dictionary, and None otherwise.


# bin(int('010101', 2))

