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 java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
import java.lang.*;
public class MAin{
	static int mod = 1000000007;
	static InputReader in = new InputReader(System.in);
	public static void main(String[] args) throws Exception {
		int t=in.readInt();
		while(t-->0) {
			int n=in.readInt();
			String s=in.readString();
			String r="1";
			char p=s.charAt(0)=='1'?'2':'1';
			for(int i=1;i<n;i++) {
				if(s.charAt(i)=='1') {
					if(p=='2') {
						r+='0';
						p='1';
					}
					else if(p=='1') {
						r+='1';
						p='2';
					}
					else if(p=='0') {
						r+=1;
						p='2';
					}
				}
				else {
					if(p=='2') {
						r+='1';
						p='1';
					}
					else if(p=='1') {
						r+='0';
						p='0';
					}
					else if(p=='0') {
						r+='1';
						p='1';
					}
				}
				
			}
			System.out.println(r);
		}
	}
	static int[]  nextIntArray(int n){
		int[] arr= new int[n];
		int i=0;
		while(i<n){
			arr[i++]=in.readInt();
		}
		return arr;
	}
	static long[] nextLongArray(int n){
		long[]arr= new long[n];
		int i=0;
		while(i<n){
			arr[i++]=in.readLong();
		}
		return arr;
	}	
	static int[]  nextIntArray1(int n) {
		int[] arr= new int[n+1]; 
		int i=1;
		while(i<=n){
			arr[i++]=in.readInt();
		}
		return arr;
	}	    	
	static long[] nextLongArray1(int n){
		long[]arr= new long[n+1];
		int i=1;
		while(i<=n) {
			arr[i++]=in.readLong();
		}
		return arr;
	}	  
	static class InputReader {
		private InputStream stream;
		private byte[] buf = new byte[1024];
		private int curChar;
		private int numChars;
		private SpaceCharFilter filter;
		public InputReader(InputStream stream) {
			this.stream = stream;
		}
		public int read() {
			if (numChars == -1)
				throw new InputMismatchException();
			if (curChar >= numChars) {
				curChar = 0;
				try {
					numChars = stream.read(buf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (numChars <= 0)
					return -1;
			}
			return buf[curChar++];
		}
		public int readInt() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			int res = 0;
			do {
				if (c < '0' || c > '9')
					throw new InputMismatchException();
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}
		public String readString() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isSpaceChar(c));
			return res.toString();
		}
		public double readDouble() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			double res = 0;
			while (!isSpaceChar(c) && c != '.') {
				if (c == 'e' || c == 'E')
					return res * Math.pow(10, readInt());
				if (c < '0' || c > '9')
					throw new InputMismatchException();
				res *= 10;
				res += c - '0';
				c = read();
			}
			if (c == '.') {
				c = read();
				double m = 1;
				while (!isSpaceChar(c)) {
					if (c == 'e' || c == 'E')
						return res * Math.pow(10, readInt());
					if (c < '0' || c > '9')
						throw new InputMismatchException();
					m /= 10;
					res += (c - '0') * m;
					c = read();
				}
			}
			return res * sgn;
		}
		public long readLong() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			long res = 0;
			do {
				if (c < '0' || c > '9')
					throw new InputMismatchException();
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}
		public boolean isSpaceChar(int c) {
			if (filter != null)
				return filter.isSpaceChar(c);
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}
		public interface SpaceCharFilter {
			public boolean isSpaceChar(int ch);
			
		}
	}
}


