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 com.sun.source.tree.Tree;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.util.TreeSet;

public class problemD {
    static class Solution {
        void solve() {
            int n = fs.nextInt();
            int[] a = fs.readArray(n);

            TreeSet[] indices = new TreeSet[n+1];
            for (int i = 0 ; i <= n ; i ++) indices[i] = new TreeSet();

            for (int i = 0 ; i < n ; i ++ ) indices[ a[i] ].add(i);

            char[] ans = new char[n];
            Arrays.fill(ans, '0');

            boolean first = true;
            for (int i = 1; i <= n ; i ++ ) if (indices[i].size() != 1) {
                first = false;
                break;
            }

            ans[n-1] = indices[1].isEmpty() ? '0' : '1';
            ans[0] = first ? '1' : '0';

            int elem = 1, s = 0, e = n -1;
            for (int k = n-2; k >= 1 ; k --) {
                int count = 0;
                count += a[s] == elem ? 1 : 0;
                count += a[e] == elem ? 1 : 0;
                if (count == 1) {
                    if (a[s] == elem) {
                        indices[elem].remove(s);
                        s++;
                    } else {
                        indices[elem].remove(e);
                        e--;
                    }
                    if (!indices[elem].isEmpty()) {
                        break;
                    }
                    elem++;
                    if (indices[elem].isEmpty()) {
                        break;
                    }
                    ans[k] = '1';
                } else {
                    break;
                }
            }

            out.println(ans);
        }
    }

    public static void main(String[] args) throws Exception {
        int T = 1;
        T = fs.nextInt();
        Solution solution = new Solution();
        for (int t = 0; t < T; t++) solution.solve();
        out.close();
    }

    static void debug(Object... O) {
        System.err.println("DEBUG: " + Arrays.deepToString(O));
    }

    private static FastScanner fs = new FastScanner();
    private static PrintWriter out = new PrintWriter(System.out);

    static class FastScanner { // Thanks SecondThread.
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");

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

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

        int[] readArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) a[i] = nextInt();
            return a;
        }

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

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

        String nextString() {
            return next();
        }
    }
}
