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. 

//package com.my.package1;
import java.util.Scanner;
import java.io.*;
import java.util.*;
public class HellWorld {

    public static void main(String[]args) throws IOException {
        int t;
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        t = Integer.parseInt(br.readLine());
        while (t > 0) {
            t--;
            int n = Integer.parseInt(br.readLine());
            String[] st = br.readLine().split(" ");
            int a[] = new int[n + 2];
            int b[] = new int[n + 2];
            for (int i = 1; i <= n; i++) {
                //a[i] = cin.nextInt();
                a[i] = Integer.parseInt(st[i - 1]);
                b[a[i]]++;
            }
            StringBuilder sb = new StringBuilder();
            int tmp = 0;
            String s = "";
            for (int i = 1; i <= n; i++) {
                if (b[i] != 1) {
                    tmp++;
                    break;
                }
            }
            int fir = 1, fin = n;
            if (tmp == 0) sb.append(1);//s += '1';
            else sb.append(0);//s += '0';
            tmp = 0;
            for (int i = 1; i <= n; i++) {
                tmp = i;
                if (b[i] != 1) break;
                if (a[fir] == i && b[i + 1] > 0 && b[i] == 1) fir++;
                else if (a[fin] == i&& b[i + 1] > 0 && b[i] == 1) fin--;
                else break;
            }
            //System.out.println(tmp);
            if(tmp == n) {
                for(int i = 2; i <= n; i++){
                    sb.append(1);
                    //System.out.print('1');
                }
                // System.out.print('\n');
                // continue;
            }

            else {

                for (int i = 2; i <= n - tmp ; i++){
                    sb.append(0);
                    //s += '0';
                }
                for (int i = n - (tmp - 1); i < n; i++) {
                    //s += '1';
                    sb.append(1);
                }
                if (b[1] > 0 && n > 1) sb.append(1);//s += '1';
                else if (n > 1) sb.append(0);//s += '0;
            }
            System.out.println(sb.toString());

        }
    }

 }
