You are given a permutation p=[p_1, p_2, …, p_n] of integers from 1 to n. Let's call the number m (1 ≤ m ≤ n) beautiful, if there exists two indices l, r (1 ≤ l ≤ r ≤ n), such that the numbers [p_l, p_{l+1}, …, p_r] is a permutation of numbers 1, 2, …, m.

For example, let p = [4, 5, 1, 3, 2, 6]. In this case, the numbers 1, 3, 5, 6 are beautiful and 2, 4 are not. It is because:

  * if l = 3 and r = 3 we will have a permutation [1] for m = 1; 
  * if l = 3 and r = 5 we will have a permutation [1, 3, 2] for m = 3; 
  * if l = 1 and r = 5 we will have a permutation [4, 5, 1, 3, 2] for m = 5; 
  * if l = 1 and r = 6 we will have a permutation [4, 5, 1, 3, 2, 6] for m = 6; 
  * it is impossible to take some l and r, such that [p_l, p_{l+1}, …, p_r] is a permutation of numbers 1, 2, …, m for m = 2 and for m = 4. 



You are given a permutation p=[p_1, p_2, …, p_n]. For all m (1 ≤ m ≤ n) determine if it is a beautiful number or not.

Input

The first line contains the only integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. The next lines contain the description of test cases.

The first line of a test case contains a number n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the given permutation p. The next line contains n integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n, all p_i are different) — the given permutation p.

It is guaranteed, that the sum of n from all test cases in the input doesn't exceed 2 ⋅ 10^5.

Output

Print t lines — the answers to test cases in the order they are given in the input. 

The answer to a test case is the string of length n, there the i-th character is equal to 1 if i is a beautiful number and is equal to 0 if i is not a beautiful number.

Example

Input


3
6
4 5 1 3 2 6
5
5 3 1 2 4
4
1 4 3 2


Output


101011
11111
1001

Note

The first test case is described in the problem statement.

In the second test case all numbers from 1 to 5 are beautiful:

  * if l = 3 and r = 3 we will have a permutation [1] for m = 1; 
  * if l = 3 and r = 4 we will have a permutation [1, 2] for m = 2; 
  * if l = 2 and r = 4 we will have a permutation [3, 1, 2] for m = 3; 
  * if l = 2 and r = 5 we will have a permutation [3, 1, 2, 4] for m = 4; 
  * if l = 1 and r = 5 we will have a permutation [5, 3, 1, 2, 4] for m = 5. 
import java.io.*;
import java.util.*;
import java.math.*;
import java.lang.*;
 
import static java.lang.Math.*;
 
public class GameTv implements Runnable 
{
    static class InputReader 
    {
        private InputStream stream;
        private byte[] buf = new byte[1024];
        private int curChar;
        private int numChars;
        private SpaceCharFilter filter;
        private BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
 
        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 String nextLine()
        {
            String str = "";
            try
            {
                str = br.readLine();
            }
            catch (IOException e)
            {
                e.printStackTrace();
            }
            return str;
        }
        public int nextInt() 
        {
            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 long nextLong() 
        {
            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 double nextDouble() 
        {
            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, nextInt());
                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, nextInt());
                    if (c < '0' || c > '9')
                        throw new InputMismatchException();
                    m /= 10;
                    res += (c - '0') * m;
                    c = read();
                }
            }
            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 boolean isSpaceChar(int c) 
        {
            if (filter != null)
                return filter.isSpaceChar(c);
            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }
     
        public String next()
        {
            return readString();
        }
        
        public interface SpaceCharFilter
        {
            public boolean isSpaceChar(int ch);
        }
    }
	
	
	// if modulo is required set value accordingly
	public static long[][] matrixMultiply2dL(long t[][],long m[][])
	{
		long res[][]= new long[t.length][m[0].length];
		
		for(int i=0;i<t.length;i++)
		{
			for(int j=0;j<m[0].length;j++)
			{
				res[i][j]=0;
				for(int k=0;k<t[0].length;k++)
				{
					res[i][j]+=t[i][k]+m[k][j];
				}
			}
		}
		
		return res;
	}
	
	static long combination(long n,long r)
	{
		long ans=1;
		for(long i=0;i<r;i++)
		{
			ans=(ans*(n-i))/(i+1);
		}
		return ans;
	}
	
	
    public static void main(String args[]) throws Exception 
    {
        new Thread(null, new GameTv(),"GameTv",1<<27).start();
    }
	// **just change the name of class from Main to reuquired**
    public void run()
    {
        InputReader sc = new InputReader(System.in);
        PrintWriter w = new PrintWriter(System.out);
		
		int tc=sc.nextInt();
		while(tc-->0)
		{
		    int pos=-1;
		    int n=sc.nextInt();
		    int a[]= new int[n];
		    for(int i=0;i<n;++i)
		    {
		        a[i]=sc.nextInt();
		        if(a[i]==1) pos=i;
		    }
		    int ans[]= new int[n+1];
		    ans[1]=1;
		    int left=pos,right=pos,max=1;
		    while(left>0 && right<n-1)
		    {
		        if(a[right+1]<a[left-1])
		        {
		            right++;
		            max=Math.max(max,a[right]);
		        }
		        else
		        {
		            left--;
		            max=Math.max(max,a[left]);
		        }
		        if((right-left+1)==max) ans[max]=1;
		    }
		    
		    if(right==n-1)
		    {
		        while(left>0)
		        {
		            left--;
		            max=Math.max(max,a[left]);
		            if((right-left+1)==max) ans[max]=1;
		        }
		    }
		    else if(left==0)
		    {
		        while(right<n-1)
		        {
		            right++;
		            max=Math.max(max,a[right]);
		            if((right-left+1)==max) ans[max]=1;
		        }
		    }
		    
		    for(int i=1;i<=n;++i) w.print(ans[i]);
		    w.println("");
		    
		    
		}
		
        System.out.flush();
        w.close();
    }
}