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 rn():
    a = int(input())
    return a


def rl():
    a = list(map(int, input().split()))
    return a

for _ in range(rn()):
    s = list(input())
    x = rn()
    n = len(s)
    a = ['2' for _ in range(n)]
    dt = []
    flag = 0
    for i in range(x):
        if i+x < n:
            a[i+x] = s[i]
        elif s[i] == '1':
            flag = 1
            break
    if flag == 0:
        for i in range(x,n-x):
            if s[i] == '1':
                dt.append([i-x,i+x])
            else:
                if a[i-x] == '1':
                    flag = 1
                    break
                else:
                    a[i-x] = '0'
                if a[i+x] == '1':
                    flag = 1
                    break
                else:
                    a[i+x] = '0'
    if flag == 0:
        for i in range(n-x,n):
            if i - x >= 0:
                if a[i-x] == '2':
                    a[i - x] = s[i]
                elif a[i-x] != s[i]:
                    flag = 1
                    break
            elif s[i] == '1':
                flag = 1
                break
    for i in dt:
        if a[i[0]] == '1' or a[i[1]] == '1':
            continue
        if a[i[0]] == '2':
            a[i[0]] = '1'
        elif a[i[1]] == '2':
            a[i[1]] = '1'
        else:
            flag = 1
            break
    if flag == 1:
        print(-1)
    else:
        for i in a:
            if i == '2':
                print('1', end = '')
            else:
                print(i, end = '')
        print('')



