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. 

/*
 * @author romit17
 */

import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.HashMap;
import java.util.InputMismatchException;

public class B1265 {
    
    void solve() throws IOException {
        PrintWriter out = new PrintWriter(System.out);
        StringBuilder sb = new StringBuilder("");
        int t = ni();
        while(t-->0)
        {
            int n = ni();
            int[] a = na(n);
            HashMap<Integer, Integer> h = new HashMap<>();
            for(int i=0;i<n;i++) h.put(a[i], i);
            int[] ans = new int[n+1];
            ans[1] = 1;
            int l,r;
            l = r = h.get(1);
            for(int i=2;i<=n;i++)
            {
                int p = h.get(i);
                if(p<l)
                {
                    if(p==l-1 && r-l+1==i-1) ans[i] = 1;
                    l = p;
                }
                else if(p>r)
                {
                    if(p==r+1 && r-l+1==i-1) ans[i] = 1;
                    r = p;
                }
                else 
                {
                    if(r-l+1==i) ans[i] = 1;
                }
            }
            for(int i=1;i<=n;i++) out.print(ans[i]);
            out.println();            
        }
        out.flush();
    }
    
    public static void main(String[] args) throws IOException {
        new B1265().solve();
    }    
    
    private byte[] inbuf = new byte[1024];
    public int lenbuf = 0, ptrbuf = 0;    
    InputStream is = System.in;

    private int readByte() {
        if (lenbuf == -1) {
            throw new InputMismatchException();
        }
        if (ptrbuf >= lenbuf) {
            ptrbuf = 0;
            try {
                lenbuf = is.read(inbuf);
            } catch (IOException e) {
                throw new InputMismatchException();
            }
            if (lenbuf <= 0) {
                return -1;
            }
        }
        return inbuf[ptrbuf++];
    }
    
    private boolean isSpaceChar(int c) {
        return !(c >= 33 && c <= 126);
    }
    
    private int skip() {
        int b;
        while ((b = readByte()) != -1 && isSpaceChar(b));
        return b;
    }
    
    private double nd() {
        return Double.parseDouble(ns());
    }
    
    private char nc() {
        return (char) skip();
    }
    
    private String ns() {
        int b = skip();
        StringBuilder sb = new StringBuilder();
        while (!(isSpaceChar(b))) { // when nextLine, (isSpaceChar(b) && b != ' ')
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }
    
    private char[] ns(int n) {
        char[] buf = new char[n];
        int b = skip(), p = 0;
        while (p < n && !(isSpaceChar(b))) {
            buf[p++] = (char) b;
            b = readByte();
        }
        return n == p ? buf : Arrays.copyOf(buf, p);
    }
    
    private char[][] nm(int n, int m) {
        char[][] map = new char[n][];
        for (int i = 0; i < n; i++) {
            map[i] = ns(m);
        }
        return map;
    }
    
    private int[] na(int n) {
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = ni();
        }
        return a;
    }
    
    private int[] na1(int n) {
        int[] a = new int[n + 1];
        for (int i = 1; i < n + 1; i++) {
            a[i] = ni();
        }
        return a;
    }
    
    private long[] nb(int n) {
        long[] a = new long[n];
        for (int i = 0; i < n; i++) {
            a[i] = nl();
        }
        return a;
    }
    
    private long[] nb1(int n) {
        long[] a = new long[n + 1];
        for (int i = 1; i < n + 1; i++) {
            a[i] = nl();
        }
        return a;
    }
    
    private int ni() {
        int num = 0, b;
        boolean minus = false;
        while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
        if (b == '-') {
            minus = true;
            b = readByte();
        }
        
        while (true) {
            if (b >= '0' && b <= '9') {
                num = num * 10 + (b - '0');
            } else {
                return minus ? -num : num;
            }
            b = readByte();
        }
    }
    
    private long nl() {
        long num = 0;
        int b;
        boolean minus = false;
        while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
        if (b == '-') {
            minus = true;
            b = readByte();
        }
        
        while (true) {
            if (b >= '0' && b <= '9') {
                num = num * 10 + (b - '0');
            } else {
                return minus ? -num : num;
            }
            b = readByte();
        }
    }
    
}

