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 sys
# sys.stdin  = open("input.txt")
for _ in range(int(input())):
    s=input()
    x=int(input())
    n=len(s)
    d=dict()
    lol=[0]*n
    lol[0]=1
    for i in range(-x,n+x):
        if i<0:
            d[i]=0
            lol[0]=1
        elif i>=n:
            d[i]=0
            lol[0]=1
        else:
            d[i]=1
            lol[0]=1
    flag=0
    for i in range(n):
        lol[0]=1
        if s[i]=="0":
            lol[0]=1
            d[i+x]=0
            lol[0]=1
            d[i-x]=0
            lol[0]=1
    for i in range(n):
        lol[0]=1
        if s[i]=="1":
            lol[0]=1
            if d[i-x]==0 and d[i+x]==0:
                lol[0]=1
                flag=1
                lol[0]=1
                break
            elif d[i-x]==0:
                lol[0]=0
                d[i+x]=1
                lol[0]=0
            else:
                lol[0]=0
                d[i-x]=1
    ans=[]
    lol[0]=0
    for i in range(n):
        lol[0]=0
        ans.append(str(d[i]))
        lol[0]=0
    lol.sort()
    print("".join(ans) if flag==0 else "-1")
