human
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
Using python3 can you solve the prior task?
assistant
#Here is Python code to print the answer on a separate line as follows:
.

#from bisect import bisect_left as bl                #c++ lowerbound bl(array,element)
#from bisect import bisect_right as br               #c++ upperbound br(array,element)
#from __future__ import print_function, division    #while using python2
# from itertools import accumulate
# from collections import defaultdict, Counter

def modinv(n,p):
    return pow(n,p-2,p)

def main():
    #sys.stdin = open('input.txt', 'r')
    #sys.stdout = open('output.txt', 'w')

    for case in range(int(input())):
        s = [int(x) for x in input()]
        x = int(input())
        n = len(s)
        w = [1] * n 
        for i in range(n):
            if s[i] == 0:
                if i - x >= 0:
                    w[i-x] = 0
                if i + x < n:
                    w[i+x] = 0
        s2 = [-1] * n
        for i in range(n):
            if i-x >= 0 and w[i-x] == 1:
                s2[i] = 1
            elif i+x < n and w[i+x] == 1:
                s2[i] = 1
            else:
                s2[i] = 0
        if s == s2:
            print(*w, sep = "")
        else:
            print(-1)
        # print(*w, sep="")
        # print(*s, sep="")
        # print(*s2, sep="")
        # print()

        

#------------------ Python 2 and 3 footer by Pajenegod and c1729-----------------------------------------
py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange

import os, sys
from io import IOBase, BytesIO

BUFSIZE = 8192
class FastIO(BytesIO):
    newlines = 0
    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None

    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
        return s
    def read(self):
        while self._fill(): pass
        return...