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 CF1450D extends PrintWriter {
	CF1450D() { super(System.out); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF1450D o = new CF1450D(); o.main(); o.flush();
	}

	void main() {
		int t = sc.nextInt();
		while (t-- > 0) {
			int n = sc.nextInt();
			int[] aa = new int[n + 2];
			aa[0] = aa[n + 1] = 0;
			for (int i = 1; i <= n; i++)
				aa[i] = sc.nextInt();
			int[] ll = new int[n + 1];
			int[] rr = new int[n + 1];
			int[] qu = new int[n + 1]; int cnt;
			cnt = 0; qu[cnt++] = 0;
			for (int i = 1; i <= n; i++) {
				int a = aa[i];
				while (aa[qu[cnt - 1]] >= a)
					cnt--;
				ll[i] = qu[cnt - 1];
				qu[cnt++] = i;
			}
			cnt = 0; qu[cnt++] = n + 1;
			for (int i = n; i >= 1; i--) {
				int a = aa[i];
				while (aa[qu[cnt - 1]] >= a)
					cnt--;
				rr[i] = qu[cnt - 1];
				qu[cnt++] = i;
			}
			int[] kk = new int[n + 1];
			for (int i = 1; i <= n; i++) {
				int a = aa[i];
				int k = rr[i] - ll[i] - 1;
				kk[a] = Math.max(kk[a], k);
			}
			int[] dd = new int[n + 2];
			for (int a = 1; a <= n; a++) {
				int k = kk[a];
				int b = n + 1 - k;
				if (a > b)
					k -= a - b;
				dd[k]++;
			}
			char[] cc = new char[n];
			for (int k = n; k >= 1; k--) {
				int d = dd[k] += dd[k + 1];
				cc[k - 1] = d == n + 1 - k ? '1' : '0';
			}
			println(cc);
		}
	}
}
