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 Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int t = Integer.valueOf(scanner.nextLine());
        for (int i = 0;i < t;i++) {
            boolean hasAns = true;
            String s = scanner.nextLine();
            int n = s.length();
            int x = Integer.valueOf(scanner.nextLine());
            int[] w = new int[n];
            for (int j = 0;j < s.length();j++) {
                if (s.charAt(j) == '0') {
                    if (j - x >= 0 && j - x < n)
                        w[j-x] = -1;
                    if (j + x >= 0 && j + x < n)
                        w[j + x] = -1;
                }
            }
            // 再来验证1的位置是否满足
            Loop1:
            for (int j = 0;j < s.length();j++) {
                if (s.charAt(j) == '1') {
                    if (j - x >= 0 && j - x < n && w[j-x] != -1) {
                        w[j-x] = 1;
                        continue;
                    }
                    if (j + x >= 0 && j + x < n && w[j+x] != -1) {
                        w[j+x] = 1;
                        continue;
                    }
                    // 不满足条件
                    hasAns = false;
                    break Loop1;
                }
            }
            if (hasAns) {
                StringBuilder sb = new StringBuilder();
                for (int j = 0;j < n;j++)
                    if (w[j] == -1)
                        sb.append("0");
                    else
                        sb.append("1");
                System.out.println(sb.toString());
            } else {
                System.out.println("-1");
            }
        }
    }
}

