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.util.*;import java.io.*;
public class Main
{

    public static void process()throws IOException
    {
        int n=ni(),arr[]=new int[n+1],res[]=new int[n+1],idx[]=new int[200001];
        //int idx_1=-1;
        for(int i=1;i<=n;i++){
            arr[i]=ni();
            
            idx[arr[i]]=i;
        }
        int sum=0;
        int l=Integer.MAX_VALUE,h=1;
        for(int i=1;i<=n;i++){
            l=Math.min(l,idx[i]);
            h=Math.max(h,idx[i]);

            if(h-l==i-1)
                p(1);
            else
                p(0);
        }
        pn("");
    }

    static FastReader sc=new FastReader();
    //static AnotherReader sc;

    public static void main(String[]args)throws IOException
    {
        //sc=new AnotherReader();
        int t=ni();
        while(t-->0)
            process();
        System.out.flush();
        System.out.close();  
    }

    static boolean multipleTC=false;
    static void pn(Object o){System.out.println(o);}
    static void p(Object o){System.out.print(o);}
    static void pni(Object o){System.out.println(o);System.out.flush();}
    static int ni()throws IOException{return sc.nextInt();}
    static long nl()throws IOException{return sc.nextLong();}
    static double nd()throws IOException{return sc.nextDouble();}
    static String nln()throws IOException{return sc.nextLine();}
    static long gcd(long a, long b)throws IOException{return (b==0)?a:gcd(b,a%b);}
    static int gcd(int a, int b)throws IOException{return (b==0)?a:gcd(b,a%b);}
    static int bit(long n)throws IOException{return (n==0)?0:(1+bit(n&(n-1)));}
//////////////////////////////////////////////////////////////////////////////////////////////////////////////////
    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 AnotherReader{BufferedReader br; StringTokenizer st;
    AnotherReader()throws FileNotFoundException{
    br = new BufferedReader(new FileReader("input.txt"));}
    String next()throws IOException{
    while (st == null || !st.hasMoreElements()) {try{
    st = new StringTokenizer(br.readLine());}
    catch (IOException  e){ e.printStackTrace(); }}
    return st.nextToken(); } int nextInt() throws IOException{
    return Integer.parseInt(next());}
    long nextLong() throws IOException
    {return Long.parseLong(next());}
    double nextDouble()throws IOException { return Double.parseDouble(next()); }
    String nextLine() throws IOException{ String str = ""; try{
    str = br.readLine();} catch (IOException e){
    e.printStackTrace();} return str;}}
/////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
}