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 Solution {
  public static void solve(BufferedReader br) throws IOException {
    int n = Integer.parseInt(br.readLine());
    int[] arr = new int[n];
    String[] s = br.readLine().split(" ");
    for (int i=0; i<n; i++) {
      arr[i] = Integer.parseInt(s[i]) - 1; // use 0-index
    }
    int[] counter = new int[n];
    for (int num : arr) {
      counter[num]++;
    }

    int[] res = new int[n];
    // we first check whether the minimum value is 1 in the array
    if (counter[0] > 0) res[n-1]=1;
    // then we check if each element occurs exactly once
    res[0] = 1;
    for (int i=0; i<n; i++) {
      if (counter[i] != 1) {
        res[0] = 0;
        break;
      }
    }

    // now we check 0 < k < n, we know the minimum value should always be at the end of interval
    // so each time we check whether the minimum value is at the end and the counter of the min value
    // is 1, if so, we know k-compression forms a permutation, and we go on until we reach some k 
    // that cannot meet the condition
    int l=0, r=n-1;
    for (int i=n-1; i>0; i--) {
      int min = n-i-1; // the minimum value in the interval [l,r], and should be at the end
      if (--counter[min] == 0 && (arr[l]==min || arr[r]==min) && counter[min+1] > 0) {
        res[i-1] = 1;
        if (arr[l] == min) l++;
        if (arr[r] == min) r--;
      } else {
        break;
      }
    }
    
    StringBuilder sb = new StringBuilder();
    for (int i:res) {
      sb.append(i);
    }
    System.out.println(sb.toString());

  }
  public static void main(String[] args) throws IOException {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    int t = Integer.parseInt(br.readLine());
    while (t-- > 0) {
      solve(br);
    }
  }
}
