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.Arrays;
import java.util.Random;
import java.util.StringTokenizer;
import java.util.TreeSet;

public class D {

    public static void main(String[] args) {
        FastReader scan = new FastReader();
        PrintWriter out = new PrintWriter(System.out);
        Task solver = new Task();
        int t = scan.nextInt();
        for(int tt = 1; tt <= t; tt++) solver.solve(tt, scan, out);
        out.close();
    }

    static class Task {

        public void solve(int testNumber, FastReader scan, PrintWriter out) {
            int n = scan.nextInt();
            TreeSet<Integer>[] pos = new TreeSet[n + 1];
            int[] minDist = new int[n + 1];
            for(int i = 0; i <= n; i++) pos[i] = new TreeSet<>();
            for(int i = 0; i < n; i++) {
                pos[scan.nextInt()].add(i);
            }
            TreeSet<Integer> have = new TreeSet<>();
            for(int i = 1; i <= n; i++) {
                if(pos[i].isEmpty()) minDist[i] = -1;
                else {
                    int bestDist = -1;
                    for(int x : pos[i]) {
                        Integer prev = have.lower(x), next = have.higher(x);
                        int total = 0;
                        total += prev == null ? x + 1 : x - prev;
                        total += next == null ? n - x : next - x;
                        bestDist = Math.max(bestDist, total - 1);
                    }
                    minDist[i] = bestDist;
                }
                have.addAll(pos[i]);
            }
            for(int i = 2; i <= n; i++) minDist[i] = Math.min(minDist[i], minDist[i - 1]);
            for(int i = 1; i <= n; i++) {
                out.print(minDist[n - i + 1] >= i ? 1 : 0);
            }
            out.println();
        }
    }

    static void ruffleSort(int[] a) {
        Random get = new Random();
        for (int i = 0; i < a.length; i++) {
            int r = get.nextInt(a.length);
            int temp = a[i];
            a[i] = a[r];
            a[r] = temp;
        }
        Arrays.sort(a);
    }

    static void ruffleSort(long[] a) {
        Random get = new Random();
        for (int i = 0; i < a.length; i++) {
            int r = get.nextInt(a.length);
            long temp = a[i];
            a[i] = a[r];
            a[r] = temp;
        }
        Arrays.sort(a);
    }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

        public FastReader(String s) throws FileNotFoundException {
            br = new BufferedReader(new FileReader(new File(s)));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

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

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

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

}