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.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.util.TreeMap;

public class ProblemG12D {
    public static void main(String[] args) throws Exception {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter writer = new PrintWriter(System.out);

        int t = Integer.parseInt(reader.readLine());

        for (int tt = 0; tt < t; tt++) {
            int n = Integer.parseInt(reader.readLine());
            int[] a = new int[n];
            StringTokenizer tokenizer = new StringTokenizer(reader.readLine());
            boolean[] hasValue = new boolean[n + 1];
            MultiTreeSet set = new MultiTreeSet();
            for (int i = 0; i < n; i++) {
                a[i] = Integer.parseInt(tokenizer.nextToken());
                hasValue[a[i]] = true;
                set.add(a[i]);
            }
            boolean permutation = true;
            for (int i = 1; i <= n; i++) {
                if (!hasValue[i]) {
                    permutation = false;
                    break;
                }
            }
            boolean[] good = new boolean[n + 1];
            if (permutation) {
                good[1] = true;
            }
            int left = 0;
            int right = n - 1;
            for (int i = 1; i <= n; i++) {
                if (set.first() == i) {
                    good[n - i + 1] = true;
                    // try remove from either end
                    if (a[left] == i) {
                        set.remove(a[left]);
                        left++;
                        continue;
                    }
                    if (a[right] == i) {
                        set.remove(a[right]);
                        right--;
                        continue;
                    }
                }
                break;
            }
            for (int i = 1; i <= n; i++) {
                writer.print(good[i] ? "1" : "0");
            }
            writer.println();
        }

        

        reader.close();
        writer.close();
    }
    static class MultiTreeSet {
        private TreeMap<Long, Long> map = new TreeMap<>();
        public void add(long x) {
            map.put(x, map.getOrDefault(x, 0L) + 1L);
        }
        public void remove(long x) {
            map.put(x, map.getOrDefault(x, 0L) - 1L);
            if (map.get(x) <= 0) {
                map.remove(x);
            }
        }
        public long first() {
            return map.firstKey();
        }
        public long last() {
            return map.lastKey();
        }
        @Override
        public String toString() {
            return map.toString();
        }
    }
}
