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. 
//Praise our lord and saviour qlf9
import java.util.*;
import java.io.*;
import java.math.*;
public class D{
public static void main(String[] omkar) throws Exception
{
   BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
   StringTokenizer st = new StringTokenizer(in.readLine());
   StringBuilder sb = new StringBuilder();
   int cases = Integer.parseInt(st.nextToken());
   for(int i = 0; i < cases; i++)
   {
      solve(in, st, sb);
   }  
   System.out.println(sb);
        }
 public static void solve(BufferedReader in, StringTokenizer st, StringBuilder sb) throws Exception
 {
  st = new StringTokenizer(in.readLine());
  int n = Integer.parseInt(st.nextToken());
  int[] bool =  new int[n];
  int[] arr = readArr(n, in, st);
  PriorityQueue<Pair> que = new PriorityQueue<Pair>();
  int[] rights = new int[n];
  int[] lefts = new int[n];
  Pair temp;
  for(int i = 0; i < n; i++)
  {
      while(que.size() > 0 && que.peek().a > arr[i])
      {
         temp = que.poll();
         rights[temp.b] = i-1;
      }
      que.add(new Pair(arr[i], i));
  }
  while(que.size() > 0)
      {
         temp = que.poll();
         rights[temp.b] = n-1;
      }
      for(int i = n-1; i >= 0; i--)
  {
      while(que.size() > 0 && que.peek().a > arr[i])
      {
         temp = que.poll();
         lefts[temp.b] = i+1;
      }
      que.add(new Pair(arr[i], i));
  }
  while(que.size() > 0)
      {
         temp = que.poll();
         lefts[temp.b] = 0;
      }
  for(int i = 0; i < n; i++)
  {
      if(arr[i] > 0 && arr[i] <= n)
      {
         bool[arr[i]-1] = Math.max(bool[arr[i]-1], rights[i]-lefts[i]+1);
      }
  }
  int min = bool[0];
  int[] ret = new int[n];
  for(int i = 0; i < n; i++)
  {
      min = Math.min(bool[i], min);
      if(min >= n-i)
      {
         ret[n-1-i] = 1;
      }
      else
      {
         ret[n-1-i] = 0;
      }
  }
  for(int i = 0; i < n; i++)
  {
   sb.append(ret[i]+"");
  }
  sb.append("\n");
 }
 public static int[] readArr(int N, BufferedReader in, StringTokenizer st) throws Exception
      {
         int[] arr = new int[N];
         st = new StringTokenizer(in.readLine());
         for(int i=0; i < N; i++)
            arr[i] = Integer.parseInt(st.nextToken());
         return arr;
      }
static class Pair implements Comparable<Pair> {
int a;
int b;
Pair(int a, int b) {
this.a = a;
this.b = b;
}
@Override
public int compareTo(Pair other) {
if (a != other.a) {
return other.a-a;
} else {
return other.b-b;
}}}


}