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
'''
    Auther: ghoshashis545 Ashis Ghosh
    College: jalpaiguri Govt Enggineering College

'''
from os import path
import sys
from heapq import heappush,heappop
from functools import cmp_to_key as ctk
from collections import deque,defaultdict as dd 
from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right
from itertools import permutations
from datetime import datetime
from math import ceil,sqrt,log,gcd
def ii():return int(input())
def si():return input().rstrip()
def mi():return map(int,input().split())
def li():return list(mi())
abc='abcdefghijklmnopqrstuvwxyz'
mod=1000000007
# mod=998244353
inf = float("inf")
vow=['a','e','i','o','u']
dx,dy=[-1,1,0,0],[0,0,1,-1]

def bo(i):
    return ord(i)-ord('a')

file = 1




    
def solve():

    

    for _ in range(ii()):

        s = si()
        x = ii()
        n = len(s)
        s1 = ['1']*n
        flag = [0]*n

        for i in range(n):
            x1 = i-x 
            x2 = i+x
            ok = False 
            if s[i]=='0':
                f = 0
                if(x1>=0):
                    f = 1
                    if(flag[x1]==0):
                        s1[x1] = '0'
                        ok = True
                    else:
                        break
                if(x2<n):
                    s1[x2]='0'
                    ok =True
                    f = 1
                if(f==0):
                    ok = True

    
            else:
                if(x1>=0 and s1[x1]=='1'):
                    ok = True
                elif(x2<n):
                    ok =True
                    flag[x2]=1
                

            if(ok == False):
                break

        if(ok == False):
            print('-1')
        else:
            print("".join(s1))
            











        
if __name__ =="__main__":

    if(file):

        if path.exists('input.txt'):
            sys.stdin=open('input.txt', 'r')
            sys.stdout=open('output.txt','w')
        else:
            input=sys.stdin.readline
    solve()
