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

t= int(input())
for _ in range(t):
    s = list(input())
    x = int(input())
    l = len(s)
    news = ['-1']*l
    flag = 1
    for i in range(l):
        #print(i)
        if s[i] == '1':
            if i-x>=0 and abs(int(news[i-x])) == 1:
                news[i-x] = '1'
                continue
            else:
                if i+x < l and abs(int(news[i+x])) == 1:
                    news[i+x] = '1'
                    continue
                else: 
                    print(-1)
                    flag = 0
                    break
        else:
            if i-x >= 0:
                if news[i-x] != '0' and news[i-x] != '-1':
                    #print("Breaking here 2")
                    print(-1)
                    flag = 0
                    break
                else:
                    news[i-x] = '0'
            
            if i+x < l:
                if news[i+x] != '0' and news[i+x] != '-1':
                    #print("Breaking here")
                    print(-1)
                    flag =0
                    break
                else:
                    news[i+x] = '0'
            elif i-x < 0:
                news[i]='0'
    #print("Out of the loop" , flag)
    if flag:
        for i in range(len(news)):
            if news[i] == '-1':
                news[i] = '1'
        print("".join(news))