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. 
import java.io.*;
import java.util.*;
public class d2prac implements Runnable 
{
	private boolean console=false;

	public void solve()
	{
		int i;
		int n=in.ni(); int a[]=new int[n]; int pos[]=new int[n];
		for(i=0;i<n;i++)
		{
			a[i]=in.ni(); pos[a[i]-1]=i;
		}
		int ans[]=new int[n]; ans[0]=1; int lo=pos[0]; int hi=pos[0];
		for(i=1;i<n;i++)
		{
			lo=Math.min(lo, pos[i]);
			hi=Math.max(hi, pos[i]);
			int dist=hi-lo;
			if(dist==i)
				ans[i]=1;
		}
		for(int v:ans)
			out.print(v);
		out.println();
	}
	@Override
	public void run() {
		try {  init();  } 
		catch (FileNotFoundException e) {  e.printStackTrace();   }

		int t= in.ni();

		while (t-->0) {
			solve();
			out.flush(); }
	}
	private FastInput in;    private PrintWriter out;
	public static void main(String[] args) throws Exception {	 new d2prac().run();	 }

	private void init() throws FileNotFoundException {
		InputStream inputStream = System.in; 	 OutputStream outputStream = System.out;
		try {  if (!console && System.getProperty("user.name").equals("sachan")) {
			outputStream = new FileOutputStream("/home/sachan/Desktop/output.txt");
			inputStream = new FileInputStream("/home/sachan/Desktop/input.txt");   	}
		}	catch (Exception ignored) {	}
		out = new PrintWriter(outputStream); 	 in = new FastInput(inputStream);
	}
	static class FastInput { InputStream obj;
	public FastInput(InputStream obj) {	this.obj = obj;	}
	byte inbuffer[] = new byte[1024];  int lenbuffer = 0, ptrbuffer = 0;
	int readByte() { if (lenbuffer == -1) throw new InputMismatchException();
	if (ptrbuffer >= lenbuffer) { ptrbuffer = 0;
	try { lenbuffer = obj.read(inbuffer);  }
	catch (IOException e) { throw new InputMismatchException(); } }    
	if (lenbuffer <= 0) return -1;return inbuffer[ptrbuffer++]; }

	String ns() { int b = skip();StringBuilder sb = new StringBuilder();
	while (!(isSpaceChar(b)))	  { sb.appendCodePoint(b);b = readByte(); }return sb.toString();}

	int ni() { int num = 0, b;boolean minus = false;
	while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ;
	if (b == '-') { minus = true;b = readByte(); }
	while (true) { if (b >= '0' && b <= '9') { num = num * 10 + (b - '0'); } else {
		return minus ? -num : num; }b = readByte(); }}

	long nl() { long num = 0;int b;boolean minus = false;
	while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ;
	if (b == '-') { minus = true;b = readByte(); }
	while (true) { if (b >= '0' && b <= '9') { num = num * 10L + (b - '0'); } else {
		return minus ? -num : num; }b = readByte(); } }

	boolean isSpaceChar(int c) { return (!(c >= 33 && c <= 126)); }
	int skip() { int b;while ((b = readByte()) != -1 && isSpaceChar(b)) ;return b; }
	float nf() {return Float.parseFloat(ns());}
	double nd() {return Double.parseDouble(ns());}
	char nc() {return (char) skip();}
	}
}
