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.
def ass(a, b):
	if a != b: print(f"Assertion error:\n{a}\n{b}")
def intline():
	return [int(item) for item in input().split()]
def line():
	return [item for item in input().split()]
def dump(lst):
	for i, item in enumerate(lst): print(i, item)
def binsearch(arr, a, b):
	m = (a + b) // 2
	if a >= b: return a - 1 if arr[a] == 0 else a
	return binsearch(arr, a, m - 1) if arr[m] == 0 else binsearch(arr, m + 1, b)

def f(b):
	arr = [int(ch) for ch in b]
	res = [1 for ch in b]
	for i,ch in enumerate(arr):
		if i == 0:
			res[i]=1+arr[i]
		else:
			key = [arr[i], res[i - 1]]
			if key==[0,0]: res[i]=1
			if key==[0,1]: res[i]=0
			if key==[0,2]: res[i]=1
			if key==[1,0]: res[i]=2
			if key==[1,1]: res[i]=2
			if key==[1,2]: res[i]=1
	return ''.join([str(res[i]-arr[i]) for i,ch in enumerate(res)])

# ass(f('0'),'1')
# ass(f('011'),'110')
# ass(f('110'),'100')
# ass(f('111000'),'101101')
# ass(f('001011'),'101110')

t = intline()[0]
for i in range(t):
	n = intline()[0]
	b = line()[0]
	print(f(b))
