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.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;
import java.util.*;

public class Code{
	static int help(int num, int[] arr, HashMap<Integer, ArrayList<Integer>> check, int[] last) {
		ArrayList<Integer> al = check.get(last[0]);
		int n = num-last[0];
		int left = al.get(0)-1;
		int right = al.get(1)+1;
		int flag=1;
		while(n>0) {
			int temp = n;
			if(left>=0&&arr[left]<=num) {
				n--;
				left--;
			}
			if(right<arr.length&&arr[right]<=num) {
				n--;
				right++;
			}
			if(temp==n) {
				flag=0;
				break;
			}
		}
		if(flag==1) {
			last[0] = num;
			al.remove(0);
			al.remove(0);
			al.add(left+1);
			al.add(right-1);
			check.put(last[0], al);
		}
		return flag;
	}
	public static void main(String[] args) throws IOException {
		BufferInput in = new BufferInput();
		StringBuilder sb = new StringBuilder();
		int t = in.nextInt();
		while(t-->0) {
			int n = in.nextInt();
			int[] arr = new int[n];
			StringBuilder ans = new StringBuilder();
			HashMap<Integer, ArrayList<Integer>> check = new HashMap<>();
			ArrayList<Integer> al = new ArrayList<>();
			int last[] = new int[1];
			last[0] = 1;
			int ind=0;
			for(int i=0;i<n;i++) {
				arr[i] = in.nextInt();
				if(arr[i]==1) ind = i;
			}
			al.add(ind);
			al.add(ind);
			check.put(1, al);
			ans.append("1");
			for(int i=2;i<=n;i++) {
				ans.append(help(i, arr, check, last));
			}
			sb.append(ans+"\n");
			
		}
		System.out.print(sb);
	}
static class BufferInput {
		
		final private int BUFFER_SIZE = 1 << 16;
		
		private DataInputStream din;
		
		private byte[] buffer;
		
		private int bufferPointer, bytesRead;
		
		public BufferInput() {
			din = new DataInputStream(System.in);
			buffer = new byte[BUFFER_SIZE];
			bufferPointer = bytesRead = 0;
		}

		public BufferInput(String file_name) throws IOException {
			din = new DataInputStream(new FileInputStream(file_name));
			buffer = new byte[BUFFER_SIZE];
			bufferPointer = bytesRead = 0;
		}

		public String readLine() throws IOException {
			byte[] buf = new byte[64];
			int cnt = 0, c;
			while ((c = read()) != -1) {
				if (c == '\n')
					break;
				buf[cnt++] = (byte) c;
			}
			return new String(buf, 0, cnt);
		}
		
		public String nextString() throws IOException{
			
			byte c = read();
			while(Character.isWhitespace(c)){
				c = read();
			}
			
			StringBuilder builder = new StringBuilder();
			builder.append((char)c);
			c = read();
			while(!Character.isWhitespace(c)){
				builder.append((char)c);
				c = read();
			}
			
			return builder.toString();
		}
		
		public int nextInt() throws IOException {
			int ret = 0;
			byte c = read();
			while (c <= ' ')
				c = read();
			boolean neg = (c == '-');
			if (neg)
				c = read();
			do {
				ret = ret * 10 + c - '0';
			} while ((c = read()) >= '0' && c <= '9');

			if (neg)
				return -ret;
			return ret;
		}
		
		public int[] nextIntArray(int n) throws IOException {
			int arr[] = new int[n];
			for(int i = 0; i < n; i++){
				arr[i] = nextInt();
			}
			return arr;
		}
		public long nextLong() throws IOException {
			long ret = 0;
			byte c = read();
			while (c <= ' ')
				c = read();
			boolean neg = (c == '-');
			if (neg)
				c = read();
			do {
				ret = ret * 10 + c - '0';
			} while ((c = read()) >= '0' && c <= '9');
			if (neg)
				return -ret;
			return ret;
		}
		
		public long[] nextLongArray(int n) throws IOException {
			long arr[] = new long[n];
			for(int i = 0; i < n; i++){
				arr[i] = nextLong();
			}
			return arr;
		}
		
		public char nextChar() throws IOException{
			byte c = read();
			while(Character.isWhitespace(c)){
				c = read();
			}
			return (char) c;	
		}
		public double nextDouble() throws IOException {
			double ret = 0, div = 1;
			byte c = read();
			while (c <= ' ')
				c = read();
			boolean neg = (c == '-');
			if (neg)
				c = read();

			do {
				ret = ret * 10 + c - '0';
			} while ((c = read()) >= '0' && c <= '9');

			if (c == '.') {
				while ((c = read()) >= '0' && c <= '9') {
					ret += (c - '0') / (div *= 10);
				}
			}

			if (neg)
				return -ret;
			return ret;
		}
		public double[] nextDoubleArray(int n) throws IOException {
			double arr[] = new double[n];
			for(int i = 0; i < n; i++){
				arr[i] = nextDouble();
			}
			return arr;
		}
		private void fillBuffer() throws IOException {
			bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
			if (bytesRead == -1)
				buffer[0] = -1;
		}
		private byte read() throws IOException {
			if (bufferPointer == bytesRead)
				fillBuffer();
			return buffer[bufferPointer++];
		}

		public void close() throws IOException {
			if (din == null)
				return;
			din.close();
		}
	}
}
