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 _1450_D {
	public static void main(String[] args) throws Exception {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter out = new PrintWriter(System.out);
		int t = Integer.parseInt(in.readLine());
		for(int i = 0; i < t; i++) {
			int n = Integer.parseInt(in.readLine());
			int[] a = new int[n];
			int[] freqs = new int[n + 1];
			StringTokenizer line = new StringTokenizer(in.readLine());
			for(int j = 0; j < n; j++) {
				a[j] = Integer.parseInt(line.nextToken());
				if(a[j] > 0 && a[j] <= n) freqs[a[j]]++;
			}
			int firstworks = freqs[1] > 0 ? 1 : 0;
			int lastworks = 1;
			for(int j = 1; j <= n; j++) {
				if(freqs[j] != 1) {
					lastworks = 0;
					break;
				}
			}
			int lpointer = 0;
			int rpointer = n - 1;
			int fail = n - 1;
			for(int j = 1; j < n; j++) {
				if(a[lpointer] == j) {
					freqs[j]--;
					lpointer++;
				}else if(a[rpointer] == j) {
					freqs[j]--;
					rpointer--;
				}else {
					fail = j;
					break;
				}
				if(freqs[j + 1] <= 0 || freqs[j] > 0) {
					fail = j;
					break;
				}
			}
			StringBuilder sb = new StringBuilder();
			if(firstworks == 0) {
				for(int j = 0; j < n; j++) {
					sb.append(0);
				}
				out.println(sb.toString());
			}else {
				sb.append(lastworks);
				for(int j = 2; j < n - fail + 1; j++) {
					sb.append(0);
				}
				for(int j = n - fail + 1; j <= n; j++) {
					sb.append(1);
				}
				out.println(sb.toString());
			}
		}
		in.close();
		out.close();
	}
}