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 javax.swing.plaf.basic.BasicScrollPaneUI;
import java.io.*;
import java.util.*;

public class Main {


    static class Reader {
        final private int BUFFER_SIZE = 1 << 16;
        private final DataInputStream din;
        private final 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[100000+1]; // 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 {
            din.close();
        }
    }

    static Reader sc = new Reader();
    static BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));


    public static void main(String[] args) throws IOException {

int t = sc.nextInt();
while(t>0){
    t--;
    int n = sc.nextInt();
    int[]  arr= new int[n+1];
    int[] index=new int[n+1];
    for(int i=1;i<=n;i++){
        arr[i]=sc.nextInt();
        index[arr[i]]=i;

    }
    boolean[] boo= new boolean[n+1];
    StringBuilder ans= new StringBuilder();
    boo[1]=true;
    ans.append(1);
    int min=index[1];
    int max=index[1];

    for(int i=2;i<=n;i++) {
        min=Math.min(min,index[i]);
        max=Math.max(max,index[i]);
        if((max-min)==(i-1)){
            ans.append(1);
        }else{
            ans.append(0);
        }

    }
    System.out.println(ans.toString());
}


     }

    private static String  printList(List<Integer> arr) {
        StringBuilder ans= new StringBuilder();
        for(int i=0;i<arr.size();i++){
            if(i!=arr.size()-1){
                ans.append(arr.get(i)).append(" ");
            }else{
                ans.append(arr.get(i));
            }
        }
        return ans.toString();

    }


    private static int gcd(int a, int b) {
        if(b==0){
            return a;
        }else{
            return gcd(b,a%b);
        }
    }


    public static int getpa ( int[] arr, int n, int k){
        int ans = 0;
        Map<Integer, Integer> map = new HashMap<>();
        ;
        for (int i = 0; i < n; i++) {
            int a = arr[i];
            if (!map.containsKey(a)) {
                map.put(a, 1);
            } else {
                int aa = map.get(a);
                aa++;
                map.put(a, aa);
            }

        }
        for (Map.Entry<Integer, Integer> entry : map.entrySet()) {
            int a = entry.getKey();
            int f = entry.getValue();
            System.out.println(a + " " + f);
            if (map.containsKey(k - a)) {
                System.out.println(map.get(k - a));
                ans += Math.max(f, map.get(k - a));
                map.put(a, 0);
            }
        }
        return ans;
    }
    public static  String printArray(int[] arr){

        StringBuilder ans= new StringBuilder();
        for(int i=0;i<arr.length;i++){
            if(i!=arr.length-1){
                ans.append(arr[i]).append(" ");
            }else{
                ans.append(arr[i]);
            }
        }
        return ans.toString();
    }
}