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 C {

	public static void main(String[] args) {
		FastScanner sc = new FastScanner();
		int t = sc.nextInt();
		StringBuilder sb = new StringBuilder();
		while(t-->0) {
			int n = sc.nextInt();
			int[] arr = new int[n];
			ArrayList<Integer>[] poss = new ArrayList[n+1];
			for(int i = 1; i <= n; i++){
				poss[i] = new ArrayList<>();
			}
			for(int i = 0; i < n; i++){
				arr[i] = sc.nextInt();
				poss[arr[i]].add(i);
			}
			TreeSet<Pair> ps = new TreeSet<>();
			ps.add(new Pair(0, n-1));
			int[] life = new int[n+1];
			for(int i = 1; i <= n; i++){
				for(int p: poss[i]) {
					Pair pa = ps.floor(new Pair(p,p));
					if(pa == null) continue;
					life[i] = Math.max(life[i], pa.b - pa.a + 1);
					ps.remove(pa);
					ps.add(new Pair(pa.a, p-1));
					ps.add(new Pair(p+1, pa.b));
				}
			}
			boolean[] valid = new boolean[n+1];
			for(int i = 1; i <= n; i++){
				if(poss[i].isEmpty()) break;
				else valid[i] = true;
			}
			int[] d = new int[n+2];
			for(int i = 1; i <= n; i++){
				if(!poss[i].isEmpty()) {
					d[0]++; d[life[i]]--;
				}
			}
			int[] dacc = new int[n];
			for(int i = 0; i < n; i++){
				if(i == 0) dacc[i] = d[i];
				else dacc[i] = dacc[i-1] + d[i];
			}
			char[] ans = new char[n];
			for(int i = 0; i < n; i++){
				if(valid[n-i] && dacc[i] == n-i) ans[i] = '1';
				else ans[i] = '0';
			}
			sb.append(ans);
			sb.append("\n");
		}
		PrintWriter pw = new PrintWriter(System.out);
		pw.println(sb.toString().trim());
		pw.flush();
	}
	
	static class Pair implements Comparable<Pair>{
		int a, b;
		public Pair(int a, int b){
			this.a = a; this.b = b;
		}
		public int compareTo(Pair p){
			return a - p.a;
		}
		public String toString(){
			return a + " " + b;
		}
	}
	
	static class FastScanner {
		public BufferedReader reader;
		public StringTokenizer tokenizer;
		public FastScanner() {
			reader = new BufferedReader(new InputStreamReader(System.in), 32768);
			tokenizer = null;
		}
		public String next() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}
		public int nextInt() {
			return Integer.parseInt(next());
		}
		public long nextLong() {
			return Long.parseLong(next());
		}
		public double nextDouble() {
			return Double.parseDouble(next());
		}
		public String nextLine() {
			try {
				return reader.readLine();
			} catch(IOException e) {
				throw new RuntimeException(e);
			}
		}
	}

}
