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. 
// package CodeForces;

import java.io.BufferedOutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.InputMismatchException;
import java.util.Stack;

public class GlobalRound12D {

	public static int[] tree;
	public static int oo = (int)1e9;
	
	public static void Build(int treeart, int end, int index, int[] ranges) {
		if(treeart == end) {
			tree[index] = ranges[treeart];
			return;
		}
		int mid = (treeart + end)>>1;
		Build(treeart, mid, index<<1, ranges);
		Build(mid + 1, end, index<<1|1, ranges);
		tree[index] = Integer.min(tree[index<<1|1], tree[index<<1]);
	}
	
	public static void Update(int treeart, int end, int index, int idx) {
		if(treeart > idx || idx > end) {
			return;
		}
		if(treeart == end) {
			tree[index] = oo;
			return;
		}
		int mid = (treeart + end)>>1;
		Update(treeart, mid, index<<1, idx);
		Update(mid + 1, end, index<<1|1, idx);
		tree[index] = Integer.min(tree[index<<1|1], tree[index<<1]);
	}
	
	public static int Query() {
		return tree[1];
	}
	
	public static void solve() {
		int t = s.nextInt();
		while(t-- > 0) {
			int n = s.nextInt();
			int[] arr = s.nextIntArray(n);
			int[] r = new int[n];
			int[] l = new int[n];
			Stack<Integer> st = new Stack<Integer>();
			st.push(0);
			for(int i = 1; i < n; i++){
				while(!st.empty() && arr[i] < arr[st.peek()]){
					int val = st.peek();
					st.pop();
					r[val] = i - 1;
				}
				st.push(i);
			}
			while(!st.empty()){
				int val = st.peek();
				st.pop();
				r[val] = n - 1;
			}
			st.push(n - 1);
			for(int i = n - 2; i >= 0; i--){
				while(!st.empty() && arr[i] < arr[st.peek()]){
					int val = st.peek();
					st.pop();
					l[val] = i + 1;
				}
				st.push(i);
			}
			while(!st.empty()){
				int val = st.peek();
				st.pop();
				l[val] = 0;
			}
			int[] ranges = new int[n + 1];
			ranges[0] = oo;
			for(int i = 0; i < n; i++) {
				ranges[arr[i]] = r[i] - l[i] + 1;
			}
			tree = new int[4 * (n + 1) + 1];
			Build(0, n, 1, ranges);
			int change = 1;
			StringBuilder ans = new StringBuilder();
			for(int i = n; i >= 1; i--) {
				int q = Query();
				if(q >= change) {
					ans.append(1);
				}else {
					ans.append(0);
				}
				Update(0, n, 1, i);
				change++;
			}
			out.println(ans);
		}
	}

	public static void main(String[] args) {
		new Thread(null, null, "Thread", 1 << 27) {
			public void run() {
				try {
					out = new PrintWriter(new BufferedOutputStream(System.out));
					s = new FastReader(System.in);
					solve();
					out.close();
				} catch (Exception e) {
					e.printStackTrace();
					System.exit(1);
				}
			}
		}.start();
	}

	public static PrintWriter out;
	public static FastReader s;

	public static class FastReader {

		private InputStream stream;
		private byte[] buf = new byte[4096];
		private int curChar, snumChars;

		public FastReader(InputStream stream) {
			this.stream = stream;
		}

		public int read() {
			if (snumChars == -1) {
				throw new InputMismatchException();
			}
			if (curChar >= snumChars) {
				curChar = 0;
				try {
					snumChars = stream.read(buf);
				} catch (IOException E) {
					throw new InputMismatchException();
				}
			}
			if (snumChars <= 0) {
				return -1;
			}
			return buf[curChar++];
		}

		public int nextInt() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			int number = 0;
			do {
				number *= 10;
				number += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return number * sgn;
		}

		public long nextLong() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			long sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			long number = 0;
			do {
				number *= 10L;
				number += (long) (c - '0');
				c = read();
			} while (!isSpaceChar(c));
			return number * sgn;
		}

		public int[] nextIntArray(int n) {
			int[] arr = new int[n];
			for (int i = 0; i < n; i++) {
				arr[i] = this.nextInt();
			}
			return arr;
		}

		public long[] nextLongArray(int n) {
			long[] arr = new long[n];
			for (int i = 0; i < n; i++) {
				arr[i] = this.nextLong();
			}
			return arr;
		}

		public String next() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isSpaceChar(c));
			return res.toString();
		}

		public String nextLine() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isEndofLine(c));
			return res.toString();
		}

		public boolean isSpaceChar(int c) {
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}

		public boolean isEndofLine(int c) {
			return c == '\n' || c == '\r' || c == -1;
		}

	}

	
}
