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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Deque;
import java.util.HashSet;
import java.util.LinkedList;

public class RatingCompression {

    static BufferedReader r = new BufferedReader(new InputStreamReader(System.in));
    static PrintWriter out = new PrintWriter(System.out);

    static int cnt[];

    public static void main(String[] args) throws IOException {
        int  t = ri();
        while(t--> 0) {
            int n = ri();
            boolean[] ans = new boolean[n];
            cnt = new int[n];
            int a[] = ria();
            int count = 0;
            for(int i = 0; i<n; i++){
                if(cnt[i]>0){
                    count++;
                }
            }
            ans[0] = count == n;
            ans[n-1] = cnt[0] > 0;
            int l = 0, r= n-1;
            for(int i = n-1; i>0; i--) {
                if(!ans[n-1]) break;
                ans[i] = true;
                int next= n - i - 1;
                if(--cnt[next] == 0 && (a[l] == next || a[r] == next) && cnt[next+1]>0) {
                    if(a[l] == next) l++;
                    if(a[r] == next)r--;
                    continue;
                }
                break;
            }
            for(boolean x: ans){
                if(x) w('1');
                else w('0');
            }
            lb();
        }
        closeOutStream();
    }

    private static boolean printSmaller(int a[], int k) {
        Deque<Integer> q = new LinkedList<>();
        int i=0;
        for(i= 0; i<k; i++){
            while(!q.isEmpty() && a[q.peekLast()] > a[i])
                q.pollLast();
            q.addLast(i);
        }
        HashSet<Integer> set = new HashSet<>();
        int maxElements = a.length - k +1;
        int thisEle;
        for(; i<a.length; ++i){
//            w(a[q.peekFirst()] +" ");
            thisEle = a[q.peekFirst()];
            if(set.contains(thisEle) || thisEle > maxElements){
                return false;
            }else {
                set.add(thisEle);
            }
            while(!q.isEmpty() && q.peekFirst() <= i-k)
                q.pollFirst();
            while(!q.isEmpty() && a[i] < a[q.peekLast()])
                q.pollLast();
            q.addLast(i);
        }
        thisEle = a[q.peekFirst()];
        if(set.contains(thisEle) || thisEle > maxElements){
            return false;
        }
        return true;
    }

    private static boolean verify(int a[], int k) {
        int l =0, r = 0;
        HashSet<Integer> set = new HashSet<>();
        while(r-l +1 <k){
            set.add(a[r++]);
        }
        for(; r<a.length; l++, r++) {
            set.add(a[r]);
            if(set.size()<k){
                return false;
            }
            set.remove(a[l]);
        }
        return true;
    }

    static int ri() throws IOException {
        return Integer.parseInt(r.readLine());
    }

    static int[] ria() throws IOException {
        String s []= r.readLine().split(" ");
        int x[] = new int[s.length];
        for(int i=0; i<s.length; i++){
            x[i] = Integer.parseInt(s[i])-1;
            cnt[x[i]]++;
        }
        return x;
    }

    static void append(String s){
        out.print(s);
    }

    static void wl(Object s){
        out.println(s);
    }

    static void w(Object c){
        out.print(c);
    }

    static void lb() {
        out.println();
    }

    static void closeOutStream(){
        out.close();
    }
}
