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 fact(x):
	if x == 0:
		return 1
	else:
		return x * fact(x-1)
def lower_bound(li, num): 
	answer = -1
	start = 0
	end = len(li)-1

	while(start <= end):
		middle = (end+start)//2
		if li[middle] >= num:
			answer = middle
			end = middle - 1
		else:
			start = middle + 1
	return answer #index where x is not less than num
def upper_bound(li, num): 
	answer = -1
	start = 0
	end = len(li)-1

	while(start <= end):
		middle = (end+start)//2

		if li[middle] <= num:
			answer = middle
			start = middle + 1
		
		else:
			end = middle - 1
	return answer #index where x is not greater than num

def abs(x):
	return x if x >=0 else -x
def binary_search(li, val, lb, ub): 
	ans = 0
	while(lb <= ub):
		mid = (lb+ub)//2
		#print(mid, li[mid])
		if li[mid] > val:
			ub = mid-1
		elif val > li[mid]:
			lb = mid + 1
		else:
			ans = 1
			break
	return ans
def kadane(x): #maximum sum contiguous subarray
	sum_so_far = 0
	current_sum = 0
	for i in x:
		current_sum += i
		if current_sum < 0:
			current_sum = 0
		else:
			sum_so_far = mpos(sum_so_far,current_sum)
	return sum_so_far
def pref(li):
	pref_sum = [0]
	for i in li:
		pref_sum.append(pref_sum[-1] + i)
	return pref_sum

#_______________________________________________________________#
'''
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'''
from math import *
for _ in range(int(input())):
#for _ in range(1):
	#n = int(input())
	#p,f = map(int,input().split())
	#s,a = map(int,input().split())
	#ws,wa = map(int,input().split())
	b = list(input())
	x = int(input())
	n = len(b)
	#s2 = input()
	#r, g, b, w = map(int,input().split())
	#a = list(map(int,input().split()))
	#b = list(map(int,input().split()))
	#s1 = list(st)
	s = [int(x) for x in b] 
	
	a = [1]*n
	for i in range(n):
		if s[i] == 0:
			if i-x >= 0:
				a[i-x] = 0
			if i+x < n:
				a[i+x] = 0
	f = 1
	for i in range(n):
		check = 0
		check = check or (i-x >= 0 and a[i-x] == 1)
		check = check or (i+x < n and a[i+x] == 1)
		if check == 0:
			if s[i] == 1:
				f = 0
				print(-1)
				break
	if f:
		print(*a,sep='')






	


















	




