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.
from collections import deque,defaultdict,Counter
import math
from sys import stdin,stdout
import sys


def inp():
	return int(stdin.readline())


def inlt():
	return list(map(int,stdin.readline().split()))


def insr():
	return list(stdin.readline().strip())


def invr():
	return map(int,stdin.readline().split())
from queue import Queue
q = Queue()
import itertools
global vis,dis,graph
def bfs(source_node,vis,graph,dis):
	q.put(source_node)
	vis[source_node]=True
	dis[source_node]=0
	while not(q.empty()):
		current_node = q.get()
		for neighbour in graph[current_node]:
			if vis[neighbour]==False:
				q.put(neighbour)
				dis[neighbour]=dis[current_node]+1
				vis[neighbour] = True
import math
def lcm(a, b):
    return abs(a*b) // math.gcd(a, b)
def main():
	for __ in range(inp()):
		n = inp()
		a = input()
		
		b = ['1' for _ in range(n)]
		bb = []
		for i in range(n):
			bb.append(int(a[i])+int(b[i]))
		for i in range(1,n):
			if bb[i]==bb[i-1]:
				b[i]='0'
				bb[i]=int(a[i])+int(b[i])
		
		print("".join(b))
		
	
			

	
	










if __name__=="__main__":
	main()


