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.lang.*;
import java.io.*;

public class B {
	private static FastReader fr;
    private static OutputStream out;
    
	private static class FastReader 
    { 
        final private int BUFFER_SIZE = 1 << 16; 
        private DataInputStream din; 
        private byte[] buffer; 
        private int bufferPointer, bytesRead; 
  
        public FastReader() 
        { 
            din = new DataInputStream(System.in); 
            buffer = new byte[BUFFER_SIZE]; 
            bufferPointer = bytesRead = 0; 
        } 
  
        public FastReader(String file_name) throws IOException 
        { 
            din = new DataInputStream(new FileInputStream(file_name)); 
            buffer = new byte[BUFFER_SIZE]; 
            bufferPointer = bytesRead = 0; 
        } 
  
        public String readLine() throws IOException 
        { 
            byte[] buf = new byte[1024];
            int cnt = 0, c; 
            while ((c = read()) != -1) 
            { 
                if (c == '\n') 
                    break; 
                buf[cnt++] = (byte) c; 
            } 
            return new String(buf, 0, cnt); 
        } 
  
        public int nextInt() throws IOException 
        { 
            int ret = 0; 
            byte c = read(); 
            while (c <= ' ') 
                c = read(); 
            boolean neg = (c == '-'); 
            if (neg) 
                c = read(); 
            do
            { 
                ret = ret * 10 + c - '0'; 
            }  while ((c = read()) >= '0' && c <= '9'); 
  
            if (neg) 
                return -ret; 
            return ret; 
        } 
  
        public long nextLong() throws IOException 
        { 
            long ret = 0; 
            byte c = read(); 
            while (c <= ' ') 
                c = read(); 
            boolean neg = (c == '-'); 
            if (neg) 
                c = read(); 
            do { 
                ret = ret * 10 + c - '0'; 
            } 
            while ((c = read()) >= '0' && c <= '9'); 
            if (neg) 
                return -ret; 
            return ret; 
        } 
  
        public double nextDouble() throws IOException 
        { 
            double ret = 0, div = 1; 
            byte c = read(); 
            while (c <= ' ') 
                c = read(); 
            boolean neg = (c == '-'); 
            if (neg) 
                c = read(); 
  
            do { 
                ret = ret * 10 + c - '0'; 
            } 
            while ((c = read()) >= '0' && c <= '9'); 
  
            if (c == '.') 
            { 
                while ((c = read()) >= '0' && c <= '9') 
                { 
                    ret += (c - '0') / (div *= 10); 
                } 
            } 
  
            if (neg) 
                return -ret; 
            return ret; 
        } 
  
        private void fillBuffer() throws IOException 
        { 
            bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE); 
            if (bytesRead == -1) 
                buffer[0] = -1; 
        } 
  
        private byte read() throws IOException 
        { 
            if (bufferPointer == bytesRead) 
                fillBuffer(); 
            return buffer[bufferPointer++]; 
        } 
  
        public void close() throws IOException 
        { 
            if (din == null) 
                return; 
            din.close(); 
        } 
        
        public int[] nextIntArray(int n) throws IOException {
        	int arr[] = new int[n];
        	for(int i=0;i<n;i++){
        		arr[i] = nextInt();
        	}
        	return arr;
        }
    } 
 
	private static void print(Object str) {
        try {
            out.write(str.toString().getBytes());
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
	
	private static void println(Object str) {
        try {
            out.write((str.toString() + "\n").getBytes());
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
	
	private static void printArray(int arr[]){
		for(int val: arr){
			print(val + " ");
		}
	}
	
	public static void main(String args[]) throws IOException{
		run();
	}
	
	private static void run() throws IOException{
		fr = new FastReader();
		out = new BufferedOutputStream(System.out);
		solve();
		out.flush();
		out.close();
	}
	
	private static void solve() throws IOException{
		int t = fr.nextInt();
		while(t-- > 0) {
			int n = fr.nextInt();
			int arr[] = new int[n];
			HashMap<Integer, Integer> map = new HashMap<>();
			for(int i=0;i<n;i++) {
				arr[i] = fr.nextInt();
				map.put(arr[i], i);
			}
			
			int ans[] = new int[n];
			int min = Integer.MAX_VALUE;
			int max = Integer.MIN_VALUE;
			
			if(map.containsKey(1)) {
				min = map.get(1);
				max = min;
				ans[0] = 1;
			}
			
			for(int i=1;i<n;i++) {
				if(map.containsKey(i+1)) {
					int idx = map.get(i+1);
					min = Math.min(min, idx);
					max = Math.max(max, idx);
				}
				
				if(max -min <= i) {
					ans[i] = 1;
				}
			}
			
			for(int i=0;i<n;i++) {
				print(ans[i]);
			}
			
			println("");
		}
	}
	
	private static int factorial(int n){
		int fac = 1;
		for(int i=2;i<=n;i++){
			fac *= i;
		}
		
		return fac;
	}
	
	private boolean isPrime(int n){
		if(n == 0 || n == 1){
			return false;
		}else if(n%2 == 0){
			return false;
		}else{
			int x = (int)(Math.sqrt(n)) + 1;
			for(int i=3;i<=x;i+=2){
				if(n % i == 0){
					return false;
				}
			}
			return true;
		}
	}
	
	private long pow(int base,int exp){
		if(exp == 0){
			return 1;
		}else if(exp == 1){
			return base;
		}else{
			long a = pow(base,exp/2);
			if(exp%2 == 0){
				return a * a;
			}else{
				return a * a * base;
			}
		}
		
	}

	private static int gcd(int a,int b){
		if(a == 0){
			return b;
		}
		
		return gcd(b%a,a);
	}
	
	private static int lcm(int a,int b){
		return (a * b)/gcd(a,b);
	}
	
	private static ArrayList<Integer> SieveOfEratosthenes(int n){
		boolean b[] = new boolean[n+1];
		b[0] = true;
		b[1] = true;
		for(int i=2;i<=5;i++){
			int j = 2;
			while(i * j <= n && !b[i]){
				b[i * j] = true;
				j++;
			}
		}
		
		ArrayList<Integer> arr = new ArrayList<>();
		for(int i=2;i<n+1;i++){
			if(!b[i]){
				arr.add(i);
			}
		}
		
		return arr;
	}

}

