Lee was cleaning his house for the party when he found a messy string under the carpets. Now he'd like to make it clean accurately and in a stylish way...

The string s he found is a binary string of length n (i. e. string consists only of 0-s and 1-s).

In one move he can choose two consecutive characters s_i and s_{i+1}, and if s_i is 1 and s_{i + 1} is 0, he can erase exactly one of them (he can choose which one to erase but he can't erase both characters simultaneously). The string shrinks after erasing.

Lee can make an arbitrary number of moves (possibly zero) and he'd like to make the string s as clean as possible. He thinks for two different strings x and y, the shorter string is cleaner, and if they are the same length, then the lexicographically smaller string is cleaner.

Now you should answer t test cases: for the i-th test case, print the cleanest possible string that Lee can get by doing some number of moves.

Small reminder: if we have two strings x and y of the same length then x is lexicographically smaller than y if there is a position i such that x_1 = y_1, x_2 = y_2,..., x_{i - 1} = y_{i - 1} and x_i < y_i.

Input

The first line contains the integer t (1 ≤ t ≤ 10^4) — the number of test cases. 

Next 2t lines contain test cases — one per two lines.

The first line of each test case contains the integer n (1 ≤ n ≤ 10^5) — the length of the string s.

The second line contains the binary string s. The string s is a string of length n which consists only of zeroes and ones.

It's guaranteed that sum of n over test cases doesn't exceed 10^5.

Output

Print t answers — one per test case.

The answer to the i-th test case is the cleanest string Lee can get after doing some number of moves (possibly zero).

Example

Input


5
10
0001111111
4
0101
8
11001101
10
1110000000
1
1


Output


0001111111
001
01
0
1

Note

In the first test case, Lee can't perform any moves.

In the second test case, Lee should erase s_2.

In the third test case, Lee can make moves, for example, in the following order: 11001101 → 1100101 → 110101 → 10101 → 1101 → 101 → 01.
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import static java.lang.Math.*;
import static java.lang.System.out;
import java.util.*;
import java.io.PrintStream;
import java.io.PrintWriter;
public class A {
 
	static final int mod = 1000000007;
	static final long temp = 998244353;
	static final long MOD = 1000000007;
	static final long M = (long)1e9+7;
 
	

	static class Pair implements Comparable<Pair>
	{
		int first, second;
		public Pair(int aa, int bb)
		{
			first = aa; second = bb;
		}
		public int compareTo(Pair p)
		{
			if(first == p.first) return (int)(second - p.second);
			return (int)(first - p.first);
		}
	}

	public static class DSU
	{
		int[] parent;
		int[] rank; //Size of the trees is used as the rank
		public DSU(int n)
		{
			parent = new int[n];
			rank = new int[n];
			Arrays.fill(parent, -1);
			Arrays.fill(rank, 1);
		}
		
		public int find(int i) //finding through path compression
		{
			return parent[i] < 0 ? i : (parent[i] = find(parent[i]));
		}
		
		public boolean union(int a, int b) //Union Find by Rank 
		{
			a = find(a);
			b = find(b);
			
			if(a == b) return false; //if they are already connected we exit by returning false.
			
			// if a's parent is less than b's parent
			if(rank[a] < rank[b])
			{
				//then move a under b
				parent[a] = b;
			}
			//else if rank of j's parent is less than i's parent
			else if(rank[a] > rank[b])
			{
				//then move b under a
				parent[b] = a;
			}
			//if both have the same rank..
			else
			{
				//move a under b (it doesnt matter if its the other way around.
				parent[b] = a;
				rank[a] = 1 + rank[a];
			}
			return true;
		}	
	}

	static class Reader {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		
		int[] readArray(int n) throws IOException {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		
		long[] longReadArray(int n) throws IOException {
			long[] a=new long[n];
			for (int i=0; i<n; i++) a[i]=nextLong();
			return a;
		}
		
		long nextLong() {
			return Long.parseLong(next());
		}
		double nextDouble() {
			return Double.parseDouble(next());
		}
	}
	
		public static boolean isPrime(int n) 
		{
			if(n == 1)
			{
				return false;
			}
			for(int i = 2;i*i<=n;i++)
			{
				if(n%i == 0)
				{
					return false;
				}
			}
			return true;
	}
	
	public static List<Integer> SieveList(int n)
	{
		boolean prime[] = new boolean[n+1];
		Arrays.fill(prime, true);
		
		List<Integer> l = new ArrayList<>();
		for (int p=2; p*p<=n; p++) 
		{ 
			if (prime[p] == true) 
			{ 	
			    for(int i=p*p; i<=n; i += p) 
				{
				    prime[i] = false; 
				}	
			} 
		} 
 
		for (int p=2; p<=n; p++) 
		{
		    if (prime[p] == true)
		    {
		       l.add(p); 
		    }
		}
		
		return l;
	}
	
	
	public static int gcd(int a, int b)
	{
		if(b == 0)
		 return a;
		
		else
		return gcd(b,a%b);
	} 
 
		public static int lcm(int a, int b)
	    {
	        return (a / gcd(a, b)) * b;
	    }
	
	public static long LongGCD(long a, long b)
	{
		if(b == 0)
			 return a;
			
			else
			return LongGCD(b,a%b);
	}
	
		public static long LongLCM(long a, long b)
	    {
	        return (a / LongGCD(a, b)) * b;
	    }
	
		public static int phi(int n)  //euler totient function
	    { 
	        int result = 1; 
	        for (int i = 2; i < n; i++) 
	            if (gcd(i, n) == 1) 
	                result++; 
	        return result; 
	    }
	
	public static long fastPow(long x, long n) 
	{
		if(n == 0)
			return 1;
		else if(n%2 == 0)
			return fastPow(x*x,n/2);
		else
			return x*fastPow(x*x,(n-1)/2);
	}
	
	   public static long modPow(long x, long y, long p)
	    {
	        long res = 1;
	        x = x % p;
	 
	        while (y > 0) {
	            if (y % 2 == 1)
	                res = (res * x) % p;
	 
	            y = y >> 1; 
	            x = (x * x) % p;
	        }
	        return res;
	    }
	   
	   static long modInverse(long n, long p)
	    {
	        return modPow(n, p - 2, p);
	    }
	 
	    // Returns nCr % p using Fermat's little theorem.
	    
	    public static long nCrModP(long n, long r,long p)
	    {
	          if (n<r) 
	              return 0;
	          
	        if (r == 0)
	            return 1;
	 
	        long[] fac = new long[(int)(n) + 1];
	        fac[0] = 1;
	 
	        for (int i = 1; i <= n; i++)
	            fac[i] = fac[i - 1] * i % p;
	 
	        return (fac[(int)(n)] * modInverse(fac[(int)(r)], p)
	                % p * modInverse(fac[(int)(n - r)], p)
	                % p)
	            % p;
	    }
	    
	    public static long nCr(long n, long r)
	    {
	    	long ans = 1;
	    	for(long i = 1,j = n;i<=r;i++,j--)
	    	{
	    		ans = ans * j / i;
	    	}
	    	return ans;
	    }
 
		public static void Sort(int[] a) {
			List<Integer> l=new ArrayList<>();
			for (int i : a) l.add(i);
			Collections.sort(l);
			//Collections.reverse(l);  //Use to Sort decreasingly 
			for (int i=0; i<a.length; i++) a[i]=l.get(i);
		}
		
		public static void ssort(char[] a)
		{
			List<Character> l = new ArrayList<>();
			for (char i : a) l.add(i);
			Collections.sort(l);
			for (int i=0; i<a.length; i++) a[i]=l.get(i);
		}
		
		
		//Modular Operations for Addition and Multiplication. 
		   public static long perfomMod(long x)
		   {
		        return ((x%M + M)%M);
		    }
		   public static long addMod(long a, long  b)
		   {
		        return perfomMod(perfomMod(a)+perfomMod(b));
		    }
		   public static long mulMod(long  a, long b)
		   {
		        return perfomMod(perfomMod(a)*perfomMod(b));
		   }
		
		   public static int LowerBound(int a[], int x) 
		   {
				  int l=-1,r=a.length;
				  while(l+1<r) {
				    int m=(l+r)>>>1;
				    if(a[m]>=x) r=m;
				    else l=m;
				  }
				  return r;
			}
		
		   
			public static int UpperBound(int a[], int x) 
			{
			    int l=-1;
			    int r=a.length;
			    while(l+1<r) {
			       int m=(l+r)>>>1;
			       if(a[m]<=x) l=m;
			       else r=m;
			    }
			    return l+1;
		}
			

	public static void main(String[] args) throws IOException 
	{
		Reader sc = new Reader();
		PrintWriter fout = new PrintWriter(System.out);
		
		int t = sc.nextInt();
		while(t-- > 0)
		{
			int n = sc.nextInt();
			char[] s = sc.next().toCharArray();
			StringBuilder app = new StringBuilder("");
			int one = -1, two = -1;
			for(int i = 0;i<n;i++)
			{
				if(s[i] == '0')
				{
					one = i;
				}
			}
			
			for(int i = n-1;i>=0;i--)
			{
				if(s[i] == '1')
				{
					two = i;
				}
			}
			
			if(one == -1 || two == -1)
			{
				app.append(s).append("\n");
			}
			else
			{
				if(one < two) app.append(s).append("\n");
				else
				{
					for(int i = 0;i<two;i++) app.append(s[i]);
					for(int i = one;i<n;i++) app.append(s[i]);
					app.append("\n");
				}
			}
			fout.println(app.toString());
		}
		
		
		fout.close();
	}
}