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. 
//package cf1;

import java.io.*;
import java.util.*;
 
public class utkarsh {
    
    BufferedReader br;
    PrintWriter out;
    
    long mod = (long) (1e9 + 7), inf = (long) (3e18);
    
    void solve() {
        int t = ni();
        while(t-- > 0) {
            int n = ni();
            int a[] = new int[n];
            int ind[] = new int[n];
            for(int i = 0; i < n; i++) {
                a[i] = ni() - 1;
                ind[a[i]] = i;
            }
            boolean vst[] = new boolean[n];
            vst[0] = true;
            int ans[] = new int[n];
            ans[0] = 1;
            int L = ind[0], R = ind[0];
            int sz = 1;
            for(int i = 1; i < n; i++) {
                if(!vst[i]) {
                    if(ind[i] < L) {
                        for(int j = ind[i]; j < L; j++) {
                            vst[a[j]] = true;   sz++;
                        }
                        L = ind[i];
                    } else if(ind[i] > R) {
                        for(int j = R+1; j <= ind[i]; j++) {
                            vst[a[j]] = true;   sz++;
                        }
                        R = ind[i];
                    }
                }
                if(sz == i+1)   ans[i] = 1;
            }
            for(int x : ans)    out.print(x);
            out.println();
        }
    }
    
    long mp(long b, long e) {
        long r = 1;
        while(e > 0) {
            if( (e&1) == 1 )    r = (r * b) % mod;
            b = (b * b) % mod;
            e >>= 1;
        }
        return r;
    }
    
    // -------- I/O Template -------------
    
    char nc() {
        return ns().charAt(0);
    }
    
    String nLine() {
        try {
            return br.readLine();
        } catch(IOException e) {
            return "-1";
        }
    }
    
    double nd() {
        return Double.parseDouble(ns());
    }
    
    long nl() {
        return Long.parseLong(ns());
    }
    
    int ni() {
        return Integer.parseInt(ns());
    }
    
    StringTokenizer ip;
    
    String ns() {
        if(ip == null || !ip.hasMoreTokens()) {
            try {
                ip = new StringTokenizer(br.readLine());
            } catch(IOException e) {
                throw new InputMismatchException();
            }
        }
        return ip.nextToken();
    }
    
    void run() {
        br = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        solve();
        out.flush();
    }
    
    public static void main(String[] args) {
        new utkarsh().run();
    }
}