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. 
/*
TASK: template
LANG: JAVA
*/
import java.io.*;
import java.lang.*;
import java.util.*;

public class D1450 {
    public static void main(String[] args) throws IOException{
        StringBuffer ans = new StringBuffer();
        StringTokenizer st;
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        st = new StringTokenizer(f.readLine());
        int t = Integer.parseInt(st.nextToken());
        for(int i = 0; i < t; i++){
            st = new StringTokenizer(f.readLine());
            int n = Integer.parseInt(st.nextToken());
            int[] arr = new int[n];
            st = new StringTokenizer(f.readLine());
            HashMap<Integer, Integer> hash = new HashMap<>();
            int minValley = Integer.MAX_VALUE;
            int smallestDupe = Integer.MAX_VALUE;
            for(int x = 0; x < n; x++){
                arr[x] = Integer.parseInt(st.nextToken());
                if(!hash.containsKey(arr[x])){
                    hash.put(arr[x],0);
                }else{
                    smallestDupe = Math.min(arr[x], smallestDupe);
                }
                hash.put(arr[x],hash.get(arr[x])+1);
                int c = x-1;
                if(c > 0 && c < n-1 && arr[c] < arr[c-1] && arr[c] < arr[c+1]){
                    minValley = Math.min(arr[c], minValley);
                }
            }
            int maxPermute = 0;
            for(int x = 1; x <= n; x++){
                if(hash.containsKey(x)){
                    maxPermute = x;
                }else
                    break;
            }
            int[] arr2 = arr.clone();
            Arrays.sort(arr2);
            int firstAppearanceValley = -1;
            int firstAppearanceDupe = -1;
            for(int x = 0; x < n; x++){
                if(arr2[x] == minValley && firstAppearanceValley == -1) {
                    firstAppearanceValley = x;
                }
                if(arr2[x] == smallestDupe && firstAppearanceDupe == -1) {
                    firstAppearanceDupe = x;
                }
            }
            int r = n-Math.min(firstAppearanceDupe, firstAppearanceValley)-1;
            if(firstAppearanceDupe == -1 && firstAppearanceValley == -1)
                r = 0;
            else if(firstAppearanceDupe == -1 || firstAppearanceValley == -1)
                r = n-1-Math.max(firstAppearanceDupe, firstAppearanceValley);
            //System.out.println(r + " " + firstAppearanceDupe + " " + firstAppearanceValley);
            for(int x = 0; x < n; x++){
                if((x >= r) && maxPermute >= n-x)
                    ans.append(1);
                else if(x == 0 && maxPermute == n)
                    ans.append(1);
                else
                    ans.append(0);

            }
            ans.append("\n");

        }
        f.close();


        System.out.println(ans);

    }
}