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.math.*;
import java.util.*;

// author : Multi-Thread

public class A {

	public static void main(String[] args) {
		int test = fs.nextInt();
//		int test = 1;
		for (int cases = 0; cases < test; cases++) {
			int n = fs.nextInt();
			char ar[] = fs.next().toCharArray();
			char ans[] = new char[n];
			if (n == 1) {
				System.out.println(1);
			} else {
				for (int i = 0; i < n;) {
					int c = 1;
					char x = ar[i];
					int j = i + 1;
					for (; j < n; j++) {
						if (ar[j] == x) {
							++c;
						} else {
							break;
						}
					}
					if (x == '1') {
						if (c % 2 == 0) {
							boolean rs = true;
							for (int k = i; k < (j < n ? j + 1 : j); k++) {
								if (rs)
									ans[k] = '1';
								else
									ans[k] = '0';
								rs = !rs;
							}
						} else {
							boolean rs = true;
							for (int k = i; k < (j < n ? j + 1 : j); k++) {
								if (rs)
									ans[k] = '1';
								else
									ans[k] = '0';
								rs = !rs;
							}
						}
					} else {
						char replace = i >= 1 ? ans[i - 1] : '1';
						for (int k = i; k < (j < n ? j + 1 : j); k++) {
							ans[k] = replace;
							replace = replace == '1' ? '0' : '1';
						}
					}
					i = j;

				}
				System.out.println(String.valueOf(ans));
			}

		}
	}

	static void intSort(int[] a, boolean reverse) {
		ArrayList<Integer> al = new ArrayList<Integer>();
		for (int i : a)
			al.add(i);
		Collections.sort(al);
		if (reverse) {
			for (int i = 0; i < a.length; i++)
				a[i] = al.get(a.length - i - 1);
		} else {
			for (int i = 0; i < a.length; i++)
				a[i] = al.get(i);
		}
	}

	static void longSort(long[] a, boolean reverse) {
		ArrayList<Long> al = new ArrayList<>();
		for (long i : al)
			al.add(i);
		Collections.sort(al);
		if (reverse) {
			for (int i = 0; i < a.length; i++)
				a[i] = al.get(a.length - i - 1);
		} else {
			for (int i = 0; i < a.length; i++)
				a[i] = al.get(i);
		}
	}

	static void sieve() {
		boolean prime[] = new boolean[101];
		Arrays.fill(prime, true);
		prime[1] = false;
		for (int i = 2; i * i <= prime.length - 1; i++) {
			for (int j = i * i; j <= prime.length - 1; j += i) {
				prime[j] = false;
			}
		}
	}

	public static int[] radixSort(int[] f) {
		return radixSort(f, f.length);
	}

	public static int[] radixSort(int[] f, int n) {
		int[] to = new int[n];
		{
			int[] b = new int[65537];
			for (int i = 0; i < n; i++)
				b[1 + (f[i] & 0xffff)]++;
			for (int i = 1; i <= 65536; i++)
				b[i] += b[i - 1];
			for (int i = 0; i < n; i++)
				to[b[f[i] & 0xffff]++] = f[i];
			int[] d = f;
			f = to;
			to = d;
		}
		{
			int[] b = new int[65537];
			for (int i = 0; i < n; i++)
				b[1 + (f[i] >>> 16)]++;
			for (int i = 1; i <= 65536; i++)
				b[i] += b[i - 1];
			for (int i = 0; i < n; i++)
				to[b[f[i] >>> 16]++] = f[i];
			int[] d = f;
			f = to;
			to = d;
		}
		return f;
	}

	static class Pair {
		int first, second;

		public Pair(int first, int second) {
			this.first = first;
			this.second = second;
		}

		public String toString() {
			return "[" + first + "," + second + "]";
		}
	}

	static class LongPair {
		long first;
		long second;

		LongPair(long a, long b) {
			this.first = a;
			this.second = b;
		}

		public String toString() {
			return "[" + first + "," + second + "]";
		}
	}

	static void primeFactors(int n) {
		while (n % 2 == 0) {
			System.out.print(2 + " ");
			n /= 2;
		}
		for (int i = 3; i <= Math.sqrt(n); i += 2) {
			while (n % i == 0) {
				System.out.print(i + " ");
				n /= i;
			}
		}
		if (n > 2)
			System.out.print(n);
	}

	static boolean isPrime(long n) {
		// Corner cases
		if (n <= 1)
			return false;
		if (n <= 3)
			return true;
		if (n % 2 == 0 || n % 3 == 0)
			return false;

		long sq = (long) Math.sqrt(n);
		for (int i = 5; i <= sq; i = i + 6)
			if (n % i == 0 || n % (i + 2) == 0)
				return false;

		return true;
	}

	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}

	static class OutputWriter {
		private final PrintWriter writer;

		public OutputWriter(OutputStream outputStream) {
			writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
		}

		public OutputWriter(Writer writer) {
			this.writer = new PrintWriter(writer);
		}

		public void print(Object... objects) {
			for (int i = 0; i < objects.length; i++) {
				if (i != 0)
					writer.print(' ');
				writer.print(objects[i]);
			}
		}

		public void println() {
			writer.print("\n");
		}

		public void printLine(Object... objects) {
			print(objects);
			writer.println();
		}

		public void close() {
			writer.close();
		}

		public void flush() {
			writer.flush();
		}

	}

	private static final FastReader fs = new FastReader();
	private static final OutputWriter op = new OutputWriter(System.out);
}