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 Global12D{
	public static void main(String[] main) throws Exception{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(br.readLine());
		PrintWriter out = new PrintWriter(System.out);
		int t = Integer.parseInt(st.nextToken());
		for(int trial = 1; trial <= t; trial++){
			st = new StringTokenizer(br.readLine());
			int n = Integer.parseInt(st.nextToken());
			int[] arr = new int[n];
			int[] freq = new int[n];
			boolean perm = true;
			st = new StringTokenizer(br.readLine());
			for(int i = 0; i < n; i++) {
				arr[i] = Integer.parseInt(st.nextToken());
				freq[arr[i]-1]++;
				if(freq[arr[i]-1] > 1)
					perm = false;
			}
			StringJoiner sj = new StringJoiner("");
			if(perm)
				sj.add("1");
			else
				sj.add("0");
			if(freq[0] > 1) {
				for(int i = 0; i < n-2; i++)
					sj.add("0");
				sj.add("1");
			}
			else if(freq[0] == 0) {
				for(int i = 0; i < n-1; i++)
					sj.add("0");
			}
			else {
				int p = 0;
				int l = 0, r = n-1;
				while(p < n && freq[p] == 1) {
					if(arr[l] == p+1) {
						l++;
						p++;
					}
					else if(arr[r] == p+1) {
						r--;
						p++;
					}
					else
						break;
				}
				if(p < n && freq[p] >= 1) {
					for(int i = 0; i < n-p-2; i++)
						sj.add("0");
					for(int i = 0; i < p+1; i++)
						sj.add("1");
				}
				else if(p < n && freq[p] == 0) {
					for(int i = 0; i < n-p-1; i++)
						sj.add("0");
					for(int i = 0; i < p; i++)
						sj.add("1");
				}
				else {
					for(int i = 0; i < p-1; i++)
						sj.add("1");
				}
			}
			out.println(sj);
		}
		out.close();
	}
}