On the competitive programming platform CodeCook, every person has a rating graph described by an array of integers a of length n. You are now updating the infrastructure, so you've created a program to compress these graphs.

The program works as follows. Given an integer parameter k, the program takes the minimum of each contiguous subarray of length k in a.

More formally, for an array a of length n and an integer k, define the k-compression array of a as an array b of length n-k+1, such that $$$b_j =min_{j≤ i≤ j+k-1}a_i$$$

For example, the 3-compression array of [1, 3, 4, 5, 2] is [min\{1, 3, 4\}, min\{3, 4, 5\}, min\{4, 5, 2\}]=[1, 3, 2].

A permutation of length m is an array consisting of m distinct integers from 1 to m in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (m=3 but there is 4 in the array).

A k-compression array will make CodeCook users happy if it will be a permutation. Given an array a, determine for all 1≤ k≤ n if CodeCook users will be happy after a k-compression of this array or not.

Input

The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.

The first line of the description of each test case contains a single integer n (1≤ n≤ 3⋅ 10^5) — the length of the array.

The second line of the description of each test case contains n integers a_1,…,a_n (1≤ a_i≤ n) — the elements of the array.

It is guaranteed, that the sum of n for all test cases does not exceed 3⋅ 10^5.

Output

For each test case, print a binary string of length n. 

The k-th character of the string should be 1 if CodeCook users will be happy after a k-compression of the array a, and 0 otherwise. 

Example

Input


5
5
1 5 3 4 2
4
1 3 2 1
5
1 3 3 3 2
10
1 2 3 4 5 6 7 8 9 10
3
3 3 2


Output


10111
0001
00111
1111111111
000

Note

In the first test case, a=[1, 5, 3, 4, 2].

  * The 1-compression of a is [1, 5, 3, 4, 2] and it is a permutation. 
  * The 2-compression of a is [1, 3, 3, 2] and it is not a permutation, since 3 appears twice. 
  * The 3-compression of a is [1, 3, 2] and it is a permutation. 
  * The 4-compression of a is [1, 2] and it is a permutation. 
  * The 5-compression of a is [1] and it is a permutation. 
import java.util.*;
import java.io.*;
import java.lang.*;
import java.math.*;
public class D {
    public static void main(String[] args) throws Exception {
        BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
        // Scanner scan = new Scanner(System.in);
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out)); 
        int tt = Integer.parseInt(bf.readLine());
        for(int t=0; t<tt; t++) {
            int n = Integer.parseInt(bf.readLine());
            StringTokenizer st = new StringTokenizer(bf.readLine());
            int[] a = new int[n]; for(int i=0; i<n; i++) a[i] = Integer.parseInt(st.nextToken());
            int counter1 = 0;
            int counter2 = n-1;
            int c = 1;
            int[] ans = new int[n];
            int[] counts = new int[n+1];
            for(int i=0; i<n; i++) counts[a[i]]++;
            if(isPermutation(a, n)) ans[n-1] = 1;
            while(counter1 <= counter2) {
                if(counts[c] > 1) break;
                if(counts[c] == 0) {
                    c -= 1;
                    break;
                }
                if(a[counter1] == c) {
                    c++;
                    counter1++;
                }
                else if(a[counter2] == c) {
                    c++;
                    counter2--;
                }
                else {
                    break;
                }
            }
            for(int i=0; i<Math.min(n,c); i++) ans[i]=1;
            StringBuilder anss = new StringBuilder();
            for(int i=n-1; i>=0; i--) {
                anss.append(ans[i]);
            }
            out.println(anss.toString());

        }
        // StringTokenizer st = new StringTokenizer(bf.readLine());
        // int[] a = new int[n]; for(int i=0; i<n; i++) a[i] = Integer.parseInt(st.nextToken());
        // int n = Integer.parseInt(st.nextToken());
        // int n = scan.nextInt();
        
        out.close(); System.exit(0);
    }
    public static boolean isPermutation(int[] a, int n) {
        boolean[] visited = new boolean[n];
        for(int i=0; i<n; i++) {
            if(a[i]-1 < n) visited[a[i]-1] = true;
        }
        for(int i=0; i<n; i++) if(!visited[i]) return false;
        return true;
    }
}
