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.*;

public class D {

	static long m = (long) (1e9 + 7);

	public static void main(String[] args) throws IOException {
		Scanner scn = new Scanner(System.in);
		StringBuilder sb = new StringBuilder();
		int T = scn.nextInt(), tcs = 0;
		D: while (tcs++ < T) {
			int n = scn.nextInt();
			int a[] = new int[n];
			for (int i = 0; i < n; i++)
				a[i] = scn.nextInt();
			int l[] = new int[n], r[] = new int[n];
			Stack<Integer> st = new Stack<>();
			for (int i = n - 1; i >= 0; --i) {
				while (!st.isEmpty() && a[st.peek()] >= a[i])
					st.pop();
				r[i] = st.isEmpty() ? n - i : st.peek() - i;
				st.push(i);
			}
			st = new Stack<>();
			for (int i = 0; i < n; i++) {
				while (!st.isEmpty() && a[st.peek()] >= a[i])
					st.pop();
				l[i] = st.isEmpty() ? i + 1 : i - st.peek();
				st.push(i);
			}
			HashMap<Integer, Integer> hm = new HashMap<>();
			for (int i = 0; i < n; i++)
				hm.put(a[i], Math.max(hm.getOrDefault(a[i], 0), l[i] + r[i] - 1));

			int minl[] = new int[n + 1];
			minl[0] = n + 2;
			for (int i = 1; i <= n; i++)
				minl[i] = Math.min(minl[i - 1], hm.getOrDefault(i, -1));

			for (int i = n, ll = 1; i >= 1; --i, ll++)
				if (minl[i] >= ll)
					sb.append(1);
				else
					sb.append(0);

			sb.append("\n");
		}
		System.out.print(sb);
	}

	static class Scanner {

		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}

		public long nextLong() throws IOException {
			return Long.parseLong(next());
		}
	}
}