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.
# include<bits/stdc++.h>

using namespace std;
# define l long long 
# define db double 
# define rep(i,a,b) for(l i=a;i<b;i++)
# define vi vector<l>
# define vvi vector<vi>
# define vsi vector<set<l> >
# define pb push_back
# define mp make_pair
# define ss second
# define ff first
# define pii pair<l,l>
# define trvi(v,it) for(vi::iterator it=v.begin();it!=v.end();++it)
# define read(a) freopen(a,"r",stdin)
# define write(a) freopen(a,"w",stdout)
# define io ios::sync_with_stdio(false)

template<typename A, typename B> ostream& operator<<(ostream &os, const pair<A, B> &p) { return os << '(' << p.first << ' ' << p.second << ')'; }

const bool MULTIPLE_TEST_CASES = true;

const l MOD=1e9+7;
const l N=1e5+5;
const l INF=1e12;

void solve() {
	l n;
	cin>>n;
	string a(n,' '),b;
	cin>>b;
	a[0] = '1';
	rep(i,1,n) {
		l prev = (a[i-1]-'0') + (b[i-1] - '0');
		l cur = b[i] - '0';
		if(prev==cur+1) {
			a[i] = '0';
		} else {
			a[i] = '1';
		}
	}
	cout<<a<<"\n";
	return;
}


int main(){
	io;
	int t=1;
	if (MULTIPLE_TEST_CASES) cin>>t;
	rep(i,0,t) {
		solve();
	}

	return 0;
}
