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.*;

public class GFG 
{
	public static void main (String[] args) 
	{
		Scanner sc = new Scanner(System.in);
		int T = sc.nextInt();
		for(int z=1 ; z<=T ; z++)
		{
		    String S = sc.next();
		    int l = S.length();
		    int x = sc.nextInt();
		    
		    char[] w = new char[l];
		    for(int i=0 ; i<l ; i++)
		       w[i] = '1';
		    
		    for(int i=0 ; i<l ; i++)
		    {
		        if(S.charAt(i)=='0')
		        {
		            if((i-x)>=0)
		              w[i-x] = '0';
		              
		            if((i+x)<l)
		              w[i+x] = '0';
		        }
		    }
		    
		    String W = new String(w);
		    
		    char[] b = new char[l];
		    for(int i=0 ; i<l ; i++)
		        b[i] = '0';
		        
		    for(int i=0 ; i<l ; i++)
		    {
		        if((i-x)>=0 && W.charAt(i-x)=='1')
		        {
		            b[i] = '1';
		        }
		        else if((i+x)<l && W.charAt(i+x)=='1')
		        {
		            b[i] = '1';
		        }
		        else
		        {
		            b[i] = '0';
		        }
		    }
		    
		    
		    String B = new String(b);
		    
		    if(S.equals(B))
		    {
		        System.out.println(W);
		    }
		    else
		    System.out.println(-1);
	
		}
	}
}