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 java.util.Arrays;
import java.util.Scanner;

public class Main {
	
	public static void main(String[] args)
	{
		Scanner s=new Scanner(System.in);
		
		StringBuilder sb=new StringBuilder();
		
		int t=s.nextInt();
		
		for(int i=0;i<t;i++)
		{
			char[] arr=s.next().toCharArray();
			
			int n=arr.length;
			
			int k=s.nextInt();
			
			int[] brr=new int[n];
			
			Arrays.fill(brr,1);
			
			for(int j=0;j<n;j++)
			{
				if(arr[j]=='0')
				{
					if(j-k>=0)
					{
						brr[j-k]=0;
					}
					if(j+k<n)
					{
						brr[j+k]=0;
					}
				}
			}
			
			int p=0;
			
			for(int j=0;j<n;j++)
			{
				if(arr[j]=='1')
				{
					if((j-k>=0&&brr[j-k]==1)||(j+k<n&&brr[j+k]==1))
					{
						
					}
					else
					{
						p=1;
						break;
					}
				}
			}
			
			if(p==0)
			{
				for(int j=0;j<n;j++)
				{
					sb.append(brr[j]);
				}
				
				sb.append("\n");
			}
			else
			{
				sb.append(-1);
				sb.append("\n");
			}
			
		}
		
		System.out.println(sb);
		
	}
	
}