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
The above is tricky. Write me a correct solution in PYTHON3.
assistant
#Here is Python code to print the answer on a separate line as follows:
.

import math
import sys
import collections

def In():
    return map(int, sys.stdin.readline().split())

input = sys.stdin.readline

def rob():
    def decide(amount, amount1, sma, big):
        def add(amount):
            nonlocal ans
            maxam = min(amount // l[pos][0], l[pos][1])
            ans += maxam
            amount -= l[pos][0] * maxam
            l[pos][1] -= maxam
            return amount

        l = [sma, big]
        pos = 0
        ans = 0
        amount = add(amount)
        if not l[pos][1]:
            pos += 1
        amount = add(amount)
        amount1 = add(amount1)
        return ans

    def solve(sma, big):
        return max(decide(p, f, sma, big), decide(f, p, sma, big))

    for _ in range(int(input())):
        p, f = In()
        sw, ax = In()
        sww, axw = In()
        if sww < axw:
            print(solve([sww, sw], [axw, ax]))
        else:
            print(solve([axw, ax], [sww, sw]))

# rob()

def binarecon():
    for i in range(int(input())):
        s = input().rstrip()
        x = int(input())
        lens = len(s)
        must = [False] * lens
        for i in range(lens):
            if s[i] == '0':
                if i - x > -1:
                    must[i - x] = True
                if i + x < lens:
                    must[i + x] = True
        for i in range(lens):
            if s[i] == '1':
                yes = False
                if i - x > -1:
                    yes = 1 - must[i - x]
                if i + x < lens:
                    yes = max(yes, 1 - must[i + x])
                if not yes:
                    print(-1)
                    break
        else:
            for i in must:
                if not i:
                    print(1,end='')
                else:
                    print(0,end='')
            print()
binarecon()