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 Main {

    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner();
        PrintWriter out = new PrintWriter(System.out);
        int tc = sc.nextInt();
        main:
        while (tc-- > 0) {
            int n = sc.nextInt();
            int[] a = sc.nextIntArray(n);
            int[] left = new int[n];
            int[] right = new int[n];
            TreeMap<Integer, Integer> map = new TreeMap<>();
            int dup = n + 1;
            for (int i = n - 1; i >= 0; i--) {
                int cur = a[i];
                Integer g = map.lowerKey(cur);
                right[i] = g != null ? map.get(g) : n;
                if (map.containsKey(cur)) dup = Math.min(cur, dup);
                map.put(cur, i);
            }
            map.clear();
            HashSet<Integer> valid = new HashSet<>();
            for (int i = 0; i < n; i++) {
                int cur = a[i];
                Integer g = map.lowerKey(cur);
                left[i] = g != null ? map.get(g) : -1;
                map.put(cur, i);
                int dist = right[i] - left[i] - 1;
                if (dist >= n - cur + 1) valid.add(cur);
            }
            int[] ans = new int[n];
            for (int x = 1; x <= Math.min(dup, n); x++) {
                if (valid.contains(x)) ans[n - x] = 1;
                else break;
            }
            if (dup > n) ans[0] = 1;
            for (int x : ans)
                out.print(x);
            out.println();
        }
        out.flush();
        out.close();
    }


    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        public Scanner(String s) throws FileNotFoundException {
            br = new BufferedReader(new FileReader(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 int[] nextIntArray(int n) throws IOException {
            int[] ans = new int[n];
            for (int i = 0; i < n; i++)
                ans[i] = nextInt();
            return ans;
        }

        public Integer[] nextIntegerArray(int n) throws IOException {
            Integer[] ans = new Integer[n];
            for (int i = 0; i < n; i++)
                ans[i] = nextInt();
            return ans;
        }

        public char nextChar() throws IOException {
            return next().charAt(0);
        }

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

        public String nextLine() throws IOException {
            return br.readLine();
        }

        public double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }

        public boolean ready() throws IOException {
            return br.ready();
        }


    }
}