In the 2022 year, Mike found two binary integers a and b of length n (both of them are written only by digits 0 and 1) that can have leading zeroes. In order not to forget them, he wanted to construct integer d in the following way: 

  * he creates an integer c as a result of bitwise summing of a and b without transferring carry, so c may have one or more 2-s. For example, the result of bitwise summing of 0110 and 1101 is 1211 or the sum of 011000 and 011000 is 022000; 
  * after that Mike replaces equal consecutive digits in c by one digit, thus getting d. In the cases above after this operation, 1211 becomes 121 and 022000 becomes 020 (so, d won't have equal consecutive digits). 



Unfortunately, Mike lost integer a before he could calculate d himself. Now, to cheer him up, you want to find any binary integer a of length n such that d will be maximum possible as integer.

Maximum possible as integer means that 102 > 21, 012 < 101, 021 = 21 and so on.

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.

The first line of each test case contains the integer n (1 ≤ n ≤ 10^5) — the length of a and b.

The second line of each test case contains binary integer b of length n. The integer b consists only of digits 0 and 1.

It is guaranteed that the total sum of n over all t test cases doesn't exceed 10^5.

Output

For each test case output one binary integer a of length n. Note, that a or b may have leading zeroes but must have the same length n.

Example

Input


5
1
0
3
011
3
110
6
111000
6
001011


Output


1
110
100
101101
101110

Note

In the first test case, b = 0 and choosing a = 1 gives d = 1 as a result.

In the second test case, b = 011 so: 

  * if you choose a = 000, c will be equal to 011, so d = 01; 
  * if you choose a = 111, c will be equal to 122, so d = 12; 
  * if you choose a = 010, you'll get d = 021. 
  * If you select a = 110, you'll get d = 121. 

We can show that answer a = 110 is optimal and d = 121 is maximum possible.

In the third test case, b = 110. If you choose a = 100, you'll get d = 210 and it's the maximum possible d.

In the fourth test case, b = 111000. If you choose a = 101101, you'll get d = 212101 and it's maximum possible d.

In the fifth test case, b = 001011. If you choose a = 101110, you'll get d = 102121 and it's maximum possible d.
import sys
testing = len(sys.argv) == 4 and sys.argv[3] == "myTest"
if testing:
    cmd = sys.stdout
    from time import time
    start_time = int(round(time() * 1000)) 
    input = open(sys.argv[1], 'r').readline
    readAll = open(sys.argv[1], 'r').read
    sys.stdout = open(sys.argv[2], 'w')
else:
    input = sys.stdin.readline
    readAll = sys.stdin.read

# ############ ---- I/O Functions ---- ############

# class InputData:
#     def __init__(self):
#         self.lines = readAll().split('\n')
#         self.n = len(self.lines)
#         self.ii = -1
#     def input(self):
#         self.ii += 1
#         assert self.ii < self.n
#         return self.lines[self.ii]
# inputData = InputData()
# input = inputData.input

def intin():
    return(int(input()))
def intlin():
    return(list(map(int,input().split())))
def chrin():
    s = input()
    return(list(s[:len(s) - 1]))
def strin():
    s = input()
    return s[:len(s) - 1]
def intlout(l, sep=" "):
    print(sep.join(map(str, l)))
    
# ############ ---- I/O Functions ---- ############

# from math import ceil
# from collections import defaultdict as dd
# from heapq import *
# from Queue import Queue

def main():
    n = intin()
    b = strin()
    if n == 1:
        return 1
    ans = ["1"]
    i = 1
    while i < n:
        prv = int(b[i-1]) + int(ans[i-1])
        if int(b[i]) + 1 != prv:
            ans.append("1")
        else:
            ans.append("0")
        i += 1
    return "".join(ans)




if __name__ == "__main__":
    for _ in xrange(intin()):
        print(main())
        # print("YES" if main() else "NO")
    # main()

    if testing:
        sys.stdout = cmd
        print(int(round(time() * 1000))  - start_time)