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
def solve(s,x):
    n=len(s)
    a=[None]*n
    todos=[]
    for i,c in enumerate(s):
        if i<=x-1:
            if i+x<n:
                if a[i+x]==None:
                    a[i+x]=int(c)
                else:
                    if int(c)!=a[i+x]: return -1
            else:
                if c=='1': return -1
        elif i>=n-x:
            if i-x>=0:
                if a[i-x]==None:
                    a[i-x]=int(c)
                else:
                    if int(c)!=a[i-x]: return -1
            else:
                if c=='1': return -1
        else:
            if c=='0':
                if a[i-x]==1 or a[i+x]==1: return -1
                a[i-x]=0
                a[i+x]=0
            else:
                todos.append((i-x,i+x))
    for i,j in todos:
        if a[i]==0 and a[j]==0: return -1
    for i in range(n):
        if a[i]==None: a[i]=1
    return ''.join(str(x) for x in a)
tn=int(input())
for ti in range(tn):
    s=input()
    x=int(input())
    ans=solve(s,x)
    print(ans)
