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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {
	public static void main(String[] args) throws NumberFormatException, IOException {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		int t = Integer.parseInt(in.readLine());
		String s;
		int x;
		test:
		for (int i = 0; i < t; i++) {
			s = in.readLine();
			x = Integer.parseInt(in.readLine());
			char[] w = new char[s.length()];
			for (int j = 0; j < s.length(); j++) {
				w[j] = '1';
			}
			for (int j = 0; j < s.length(); j++) {
				if (s.charAt(j) == '0') {
					if (j-x >= 0)
						w[j-x] = '0';
					if (j+x < s.length())
						w[j+x] = '0';
				}
			}
			for (int j = 0; j < s.length(); j++) {
				if (s.charAt(j) == '1') {
					if ((j-x < 0 || w[j-x] == '0') && (j+x >= s.length() || w[j+x] == '0')) {
						System.out.println(-1);
						continue test;
					}
				}
			}
			System.out.println(w);
		}
	}
}
