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.*;

public class Main {
    static Scanner sc = new Scanner(System.in);
    static PrintWriter out = new PrintWriter(System.out);
    static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    //    long mod = 998244353;
    long mod = (long)1e9+7;
    public static void main(String[] args) throws Exception {
        Main main = new Main();
        int t = sc.nextInt();
//        int t= 1;
        /*
        while(t-->0){
            main.solve();
        }*/
        for(int i=1; i<=t; i++){
            main.solve();
        }
        out.flush();
    }

    void solve() throws Exception {
        int n = neInt();
        int[] a = new int[n+2];
        for(int i=1; i<=n; i++) a[i] = neInt();
        a[0] = -1; a[n+1] = -1;
        int[] prev = new int[n+2], suf = new int[n+2];
        Stack<Integer> st = new Stack<>();
        // front to end
        st.push(0);
        for(int i=1; i<=n; i++){
            while(st.size()>0&&a[st.peek()]>=a[i]) st.pop();
            prev[i] = i-st.peek();
            st.push(i);
        }
        st.clear();
        st.push(n+1);
        for(int i=n; i>0; i--){
            while(st.size()>0&&a[st.peek()]>=a[i]) st.pop();
            suf[i] = st.peek()-i;
            st.push(i);
        }
        List<Integer>[] rec = new List[n+1];
        for(int i=0; i<=n; i++) rec[i] = new LinkedList<>();
        for(int i=1; i<=n; i++){
            int v = a[i], len = prev[i]+suf[i]-1;
            rec[v].add(len);
        }
        StringBuilder sb = new StringBuilder();
        int mini = Integer.MAX_VALUE/2;
        for(int v=1; v<=n; v++){
            int tempMax=0;
            for(int i:rec[v]) tempMax = Math.max(tempMax, i);
            mini = Math.min(mini, tempMax);
            if(mini<n+1-v) sb.append(0);
            else sb.append(1);
        }
        out.println(sb.reverse().toString());
    }
    String neS(){
        return sc.next();
    }
    int neInt(){
        return Integer.parseInt(sc.next());
    }
    long neLong(){
        return Long.parseLong(sc.next());
    }

    int paIn(String s){return Integer.parseInt(s);}

}