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.*;
import java.util.*;
import java.math.*;

public class D {
	static byte[] buf = new byte[1<<26];
    static int bp = -1;	
	
	public static void main(String[] args) throws IOException {

		/**/
		DataInputStream in = new DataInputStream(System.in);
		/*/
		DataInputStream in = new DataInputStream(new FileInputStream("src/d.in"));
		/**/
		
		in.read(buf, 0, 1<<26);
		int t = nni();
		for (int z = 0; z < t; ++z) {
			int n = nni();
			int[] a = new int[n];
			for (int i = 0; i < n; ++i)
				a[i] = nni();
			int[] cts = new int[n+1];
			for (int x : a)
				cts[x]++;
			boolean[] hap = new boolean[n+1];
			boolean[] unhap = new boolean[n+1];
			TreeSet<Integer> rem = new TreeSet<Integer>();
			int[] some = new int[n+1];
			for (int i = 0; i < n; ++i) {
				rem.add(i);
				some[a[i]] = i;
			}
			for (int i = 1; i <= n; ++i) {
				if (!unhap[i]&&cts[i]>=1)
					hap[i] = true;
				if (a[rem.first()]!=i&&a[rem.last()]!=i)
					break;
				if (cts[i]!=1)
					break;
				rem.remove(some[i]);
			}
			hap[n] = true;
			for (int i = 1; i <= n; ++i)
				if (cts[i]!=1)
					hap[n] = false;
			StringBuilder ans = new StringBuilder();
			for (int i = n; i>=1; --i)
				ans.append(hap[i]?'1':'0');
			System.out.println(ans);
		}
	}
		
	public static int nni() {
        int ret = 0;
        byte b = buf[++bp];
        while (true) {
            ret = ret*10+b-'0';
            b = buf[++bp];
            if (b<'0'||b>'9') {
            	while (buf[bp+1]=='\r'||buf[bp+1]=='\n'||buf[bp+1]==' ') {++bp;}
            	break;
            }
        }
        return ret;
    }
}