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. 

 //   * * * its fun to do the impossible * * *   //
 import java.io.BufferedReader;
 import java.io.IOException;
 import java.io.InputStreamReader;
 import java.util.*;
  
 
 public class BeautifulNumbers {
     public static void main(String[] args) {
        FastScanner sc = new FastScanner();
        int t = sc.nextInt();

        while(t-->0){
            int n = sc.nextInt();
            int p[] = new int[n];
            for(int i = 0; i < n; i++){
                int x = sc.nextInt();
                p[x - 1] = i;
            }

            int res[] = new int[n];

            int l = n;
            int r = 0;

            for(int i = 0; i < n; i++){
                l = Math.min(p[i] , l);
                r = Math.max(p[i] , r);

                if(r - l == i){
                    res[i] = 1;
                }
            }

            for(int e : res){
                System.out.print(e);
            }

            System.out.println();
        }
     }
 
 
 
 
 
 
  
     // Use this instead of Arrays.sort() on an array of ints. Arrays.sort() is n^2
     // worst case since it uses a version of quicksort. Although this would never
     // actually show up in the real world, in codeforces, people can hack, so
     // this is needed.
     static void ruffleSort(int[] a) {
         //ruffle
         int n=a.length;
         Random r=new Random();
         for (int i=0; i<a.length; i++) {
             int oi=r.nextInt(n), temp=a[i];
             a[i]=a[oi];
             a[oi]=temp;
         }
         
         //then sort
         Arrays.sort(a);
     }
     
     // Use this to input code since it is faster than a Scanner
     static class FastScanner {
         BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
         StringTokenizer st=new StringTokenizer("");
         String next() {
             while (!st.hasMoreTokens())
                 try {
                     st=new StringTokenizer(br.readLine());
                 } catch (IOException e) {
                     e.printStackTrace();
                 }
             return st.nextToken();
         }
         
         int nextInt() {
             return Integer.parseInt(next());
         }
 
         double nextDouble() {
             return Double.parseDouble(next());
         }
  
         String str = "";
  
         String nextLine() {
             try {
                 str = br.readLine();
             } catch (IOException e) {
                 e.printStackTrace();
             }
             return str;
         }
  
         int[] readArray(int n) {
             int[] a = new int[n];
             for (int i = 0; i < n; i++)
                 a[i] = nextInt();
             return a;
         }
  
         long[] readLongArray(int n) {
             long[] a = new long[n];
             for (int i = 0; i < n; i++)
                 a[i] = nextLong();
             return a;
         }
         long nextLong() {
             return Long.parseLong(next());
         }
     }
     // use this to find the index of any element in the array +1 /// 
     // returns an array that corresponds to the index of the i+1th in the array a[]
     // runs only for array containing different values enclosed btw 0 to n-1
     static int[] indexOf(int[] a) {
         int[] toRet=new int[a.length];
         for (int i=0; i<a.length; i++) {
             toRet[a[i]]=i+1;
         }
         return toRet;
     }
 
     static int gcd(int a, int b) {
         if (b==0) return a;
         return gcd(b, a%b);
     }
 
     //generates all the prime numbers upto n
     static void sieveOfEratosthenes(int n , ArrayList<Integer> al)
     {
         boolean prime[] = new boolean[n + 1];
         for (int i = 0; i <= n; i++)
             prime[i] = true;
  
         for (int p = 2; p * p <= n; p++) 
         {
             if (prime[p] == true) 
             {
                 for (int i = p * p; i <= n; i += p)
                     prime[i] = false;
             }
         }
         for (int i = 2; i <= n; i++)
         {
             if (prime[i] == true)
                 al.get(i);
         }
     }
     static final int mod=100000000 + 7;
     static final int max_val = 2147483647;
     static final int min_val = max_val + 1;
     
     //fastPow
     static long fastPow(long base, long exp) {
         if (exp==0) return 1;
         long half=fastPow(base, exp/2);
         if (exp%2==0) return mul(half, half);
         return mul(half, mul(half, base));
     }
 
     //multiply two long numbers
     static long mul(long a, long b) {
         return a*b%mod;
     }
 
     static int nCr(int n, int r)
     {
            return fact(n) / (fact(r) *
                 fact(n - r));
     }
     
     // Returns factorial of n
     static int fact(int n)
     {
         int res = 1;
         for (int i = 2; i <= n; i++)
             res = res * i;
         return res;
     }
      
     // to generate the lps array
     // lps means longest preffix that is also a suffix
     static void generateLPS(int lps[] , String p){
         int l = 0;
         int r = 1;
    
         while(l < p.length() && l < r && r < p.length()){
             if(p.charAt(l) == p.charAt(r)){
                 lps[r] = l + 1;
                 l++;
                 r++;
             }
             else{
                 if(l > 0)
                     l = lps[l - 1];
                 else
                     r++;    
             }
         }
     }
      
 
     //count sort --> it runs in O(n) time but compromises in space
     static ArrayList<Integer> countSort(int a[]){
         int max = Integer.MIN_VALUE;
         for(int i=0;i<a.length;i++){
             max = Math.max(max , a[i]);
         }
 
         int posfre[] = new int[max+1];
         boolean negPres = false;
         for(int i=0;i<a.length;i++){
             if(a[i]>=0){
                 posfre[a[i]]++;
             }
             else{
                 negPres = true;
             }
         }
         ArrayList<Integer> res = new ArrayList<>();
         if(negPres){
             int min = Integer.MAX_VALUE;
             for(int i=0;i<a.length;i++){
                 min = Math.min(min , a[i]);
             }
 
             int negfre[] = new int[-1*min+1];
             for(int i=0;i<a.length;i++){
                 if(a[i]<0){
                     negfre[-1*a[i]]++;
                 }
             }
 
             for(int i=min;i<0;i++){
                 for(int j=0;j<negfre[-1*i];j++){
                     res.add(i);
                 }
             }
             for(int i=0;i<=max;i++){
                 for(int j=0;j<posfre[i];j++){
                     res.add(i);
                 }
             }
             return res;
         }
         for(int i=0;i<=max;i++){
             for(int j=0;j<posfre[i];j++){
                 res.add(i);
             }
         }
         return res;
     }
 
     // returns the index of the element which is just smaller than or
     // equal to the tar in the given arraylist 
     static int lowBound(ArrayList<Integer> ll,long tar ,int l,int r){
         if(l>r) return l;
 
         int mid=l+(r-l)/2;
 
         if(ll.get(mid)>=tar){
             return lowBound(ll,tar,l,mid-1);
         }
 
         return lowBound(ll,tar,mid+1,r);
     }
 
     // returns the index of the element which is just greater than or
     // equal to the tar in the given arraylist 
     static int upBound(ArrayList<Integer> ll,long tar,int l ,int r){
         if(l>r) return l;
 
         int mid=l+(r-l)/2;
 
         if(ll.get(mid)<=tar){
             return upBound(ll,tar,l,mid-1);
         }
 
         return upBound(ll,tar,mid+1 ,r);
     }
 
     // a -> z == 97 -> 122
 
     // String.format("%.9f", ans) ,--> to get upto 9 decimal places , (ans is double)
     
     // write 
 }
 