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.*;

public class Main { 
    
    public static void main(String[] args) throws java.lang.Exception { 
        Reader pm =new Reader(); 
        //Scanner pm = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        int t = pm.nextInt();
        while(t-- > 0){
            int n = pm.nextInt();
            int[] a = new int[n];
            HashMap<Integer, Integer> hm = new HashMap<>();
            for(int i=0;i<n;i++) {
                a[i] = pm.nextInt();
                hm.put(a[i], i+1);
            }
            StringBuffer sb = new StringBuffer();
            sb.append("1");
            if(n == 1) {
                System.out.println(sb);
                continue;
            }
            TreeSet<Integer> ts = new TreeSet<>();
            ts.add(hm.get(1));
            for(int i=2;i<=n;i++){
                int ind = hm.get(i);
                ts.add(ind);
                if((ts.last() - ts.first() + 1) == ts.size()) 
                    sb.append("1");
                else
                    sb.append("0");
            }
            System.out.println(sb);
        }
        //end of tests
        
    }
    //end of main class

    static int countInRange(int arr[], int n, int x, int y) { 
        // initialize result 
        int count = 0; 
        count = upperIndex(arr, n, y) -  
                lowerIndex(arr, n, x) + 1; 
        return count; 
    } 
    
    
    static int lowerIndex(int arr[], int n, int x) { 
        int l = 0, h = n - 1; 
        while (l <= h)  
        { 
            int mid = (l + h) / 2; 
            if (arr[mid] >= x) 
                h = mid - 1; 
            else
                l = mid + 1; 
        } 
        return l; 
    } 
      
    // function to find last index <= y 
    static int upperIndex(int arr[], int n, int y) { 
        int l = 0, h = n - 1; 
        while (l <= h)  
        { 
            int mid = (l + h) / 2; 
            if (arr[mid] <= y) 
                l = mid + 1; 
            else
                h = mid - 1; 
        } 
        return h; 
    }
    
    public static StringBuilder dec_to_bin(long n) {
        // decimal to binary upto 30 binaries
        if(n==0) {
            StringBuilder str=new StringBuilder("");
            for(int i=0;i<30;i++) {
                str.append("0");
            }
            return str;
        }
        StringBuilder str=new StringBuilder("");
        while(n!=0) {
            str.append(n%2L);
            n/=2L;
        }
        str=str.reverse();
        StringBuilder tmp_str=new StringBuilder("");
        int len=str.length();
        while(len!=30) {
            tmp_str.append("0");
            len++;
        }
        tmp_str.append(str);
        return tmp_str;
    }
    
    private static int binarySearchPM(int[] arr, int key){
        int n = arr.length;
        int mid = -1;
        int begin = 0,end=n;
        while(begin <= end){
            mid = (begin+end) / 2;
            if(mid == n){
                return n;
            }
            if(key < arr[mid]){
                end = mid-1;
            } else if(key > arr[mid]){
                begin = mid+1;
            } else {
                return mid;
            }
        }
        //System.out.println(begin+" "+end);
        return -begin; //expected Index
    }
    
    static class Reader { 
        final private int BUFFER_SIZE = 1 << 16; 
        private DataInputStream din; 
        private byte[] buffer; 
        private int bufferPointer, bytesRead; 
  
        public Reader() 
        { 
            din = new DataInputStream(System.in); 
            buffer = new byte[BUFFER_SIZE]; 
            bufferPointer = bytesRead = 0; 
        } 
  
        public Reader(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[64]; // line length 
            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(); 
        } 
    }
}    