You are given a permutation p=[p_1, p_2, …, p_n] of integers from 1 to n. Let's call the number m (1 ≤ m ≤ n) beautiful, if there exists two indices l, r (1 ≤ l ≤ r ≤ n), such that the numbers [p_l, p_{l+1}, …, p_r] is a permutation of numbers 1, 2, …, m.

For example, let p = [4, 5, 1, 3, 2, 6]. In this case, the numbers 1, 3, 5, 6 are beautiful and 2, 4 are not. It is because:

  * if l = 3 and r = 3 we will have a permutation [1] for m = 1; 
  * if l = 3 and r = 5 we will have a permutation [1, 3, 2] for m = 3; 
  * if l = 1 and r = 5 we will have a permutation [4, 5, 1, 3, 2] for m = 5; 
  * if l = 1 and r = 6 we will have a permutation [4, 5, 1, 3, 2, 6] for m = 6; 
  * it is impossible to take some l and r, such that [p_l, p_{l+1}, …, p_r] is a permutation of numbers 1, 2, …, m for m = 2 and for m = 4. 



You are given a permutation p=[p_1, p_2, …, p_n]. For all m (1 ≤ m ≤ n) determine if it is a beautiful number or not.

Input

The first line contains the only integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. The next lines contain the description of test cases.

The first line of a test case contains a number n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the given permutation p. The next line contains n integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n, all p_i are different) — the given permutation p.

It is guaranteed, that the sum of n from all test cases in the input doesn't exceed 2 ⋅ 10^5.

Output

Print t lines — the answers to test cases in the order they are given in the input. 

The answer to a test case is the string of length n, there the i-th character is equal to 1 if i is a beautiful number and is equal to 0 if i is not a beautiful number.

Example

Input


3
6
4 5 1 3 2 6
5
5 3 1 2 4
4
1 4 3 2


Output


101011
11111
1001

Note

The first test case is described in the problem statement.

In the second test case all numbers from 1 to 5 are beautiful:

  * if l = 3 and r = 3 we will have a permutation [1] for m = 1; 
  * if l = 3 and r = 4 we will have a permutation [1, 2] for m = 2; 
  * if l = 2 and r = 4 we will have a permutation [3, 1, 2] for m = 3; 
  * if l = 2 and r = 5 we will have a permutation [3, 1, 2, 4] for m = 4; 
  * if l = 1 and r = 5 we will have a permutation [5, 3, 1, 2, 4] for m = 5. 
import java.util.*;
import java.io.*;
import java.math.BigInteger;

public class tr0 {
	static PrintWriter out;
	static StringBuilder sb;
	static final double EPS = 1e-9;
	static long mod = (int) 1e9 + 7;
	static int inf = (int) 1e9 + 2;
	static long[] fac, a;
	static int[] si;
	static ArrayList<Integer> primes;
	static ArrayList<Integer>[] ad;
	static ArrayList<pair>[] d;
	static edge[] ed;
	static int[][] vis;
	static int[] l, ch;
	static int[] occ;
	static Queue<Integer>[] can;
	static String s;
	static long[][] memo;
	static int n;

	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		out = new PrintWriter(System.out);
		int t = sc.nextInt();
		while (t-- > 0) {
			int n=sc.nextInt();
			qu []a=new qu [n];
			for(int i=0;i<n;i++)
				a[i]=new qu(sc.nextInt()-1, i);
			Arrays.sort(a);
			boolean []e=new boolean [n];
			int er=1;
		  sb=new StringBuilder();
		  sb.append('1');
		  if(n==1) {
			  out.println(1);
			  continue;
		  }
		//  unionfind uf=new unionfind(n+1);
		    e[a[0].val]=true;
		    int []ta=new int [n];
		    ta[a[0].val]=1;
			for(int i=1;i<n;i++) {
			         int id=a[i].val;
			         ta[id]=a[i].id;
			         if(id==0) {
			        	 if(ta[id+1]==0) {
			        		 er++;
			        	 }
			         }
			         else if(id==n-1) {
			        	 if(ta[id-1]==0) {
			        		 er++;
			        	 }
			         }
			         else {
			        	 if(ta[id+1]!=0 && ta[id-1]!=0) {
			        		 er--;
			        	 }
			        	 if(ta[id+1]==0 && ta[id-1]==0) {
			        		 er++;
			        	 }
			         }
			         
			         if(er==1) 
			        	 sb.append('1');
			        	 else
			        		 sb.append('0');
			 }
          out.println(sb);
		}

		out.flush();
	}

	static class qu implements Comparable<qu> {
		int id, val;

		qu(int i, int v) {
			id = i;
			val = v;
		}

		public String toString() {
			return id + " " + val;
		}

		public int compareTo(qu o) {
			return id - o.id;
		}
	}

	static class seg implements Comparable<seg> {
		int a;
		int b;

		seg(int s, int e) {
			a = s;
			b = e;
		}

		public String toString() {
			return a + " " + b;
		}

		public int compareTo(seg o) {
			// if(a==o.a)
			return o.b - b;
			// return
		}
	}

	static class pair {
		int to;
		int number;

		pair(int t, int n) {
			number = n;
			to = t;
		}

		public String toString() {
			return to + " " + number;
		}
	}

	static double modPow(double a, int e)

	{
		double res = 1;
		while (e > 0) {
			if ((e & 1) == 1)
				res = (res * a);
			a = (a * a);
			e >>= 1;
		}
		return res;
	}

	static boolean[] in;

	/*
	 * static void mst() { Arrays.sort(ed); UnionFind uf=new UnionFind(n); for(int
	 * i=0;i<m;i++) { edge w=ed[i]; if(!uf.union(w.from, w.to)) continue;
	 * in[i]=true; } }
	 */
	static class edge implements Comparable<edge> {
		int from;
		int to;
		int number;

		edge(int f, int t, int n) {
			from = f;
			to = t;
			number = n;
		}

		public String toString() {
			return from + " " + to + " " + number;
		}

		public int compareTo(edge f) {
			return f.number - number;
		}
	}

	static long power(int i) {
		// if(i==0)
		// return 1;
		long a = 1;
		for (int k = 0; k < i; k++)
			a *= i;
		return a;
	}

	static void seive(int N) {
		si = new int[N];
		primes = new ArrayList<>();
		si[1] = 1;
		for (int i = 2; i < N; i++) {
			if (si[i] == 0) {
				si[i] = i;
				primes.add(i);
			}
			for (int j = 0; j < primes.size() && primes.get(j) <= si[i] && (i * primes.get(j)) < N; j++)
				si[primes.get(j) * i] = primes.get(j);

		}
	}

	static long fac(int n) {
		if (n == 0)
			return fac[n] = 1;
		if (n == 1)
			return fac[n] = 1;
		long ans = 1;
		for (int i = 1; i <= n; i++)
			fac[i] = ans = (i % mod * ans % mod) % mod;
		return ans % mod;
	}

	static long gcd(long a, long b) {

		if (b == 0)
			return a;
		return gcd(b, a % b);
	}

	static class unionfind {
		int[] p;
		int[] size;
        int num;
		unionfind(int n) {
			p = new int[n];
			size = new int[n];

			for (int i = 0; i < n; i++) {
				p[i] = i;
			}
			Arrays.fill(size, 1);
			num=0;
		}

		int findSet(int v) {
			if (v == p[v])
				return v;
			return p[v] = findSet(p[v]);
		}

		boolean sameSet(int a, int b) {
			a = findSet(a);
			b = findSet(b);
			if (a == b)
				return true;
			return false;
		}

		int max() {
			int max = 0;
			for (int i = 0; i < size.length; i++)
				if (size[i] > max)
					max = size[i];
			return max;
		}

		boolean combine(int a, int b) {
			a = findSet(a);
			b = findSet(b);
			if (a == b)
				return true;
			if (size[a] > size[b]) {
				p[b] = a;
				size[a] += size[b];

			} else {
				p[a] = b;
				size[b] += size[a];
			}
			return false;
		}
	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}

		public Scanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public String nextLine() throws IOException {
			return br.readLine();
		}

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

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

		public char nextChar() throws IOException {
			return next().charAt(0);
		}

		public Long nextLong() throws IOException {
			return Long.parseLong(next());
		}

		public boolean ready() throws IOException {
			return br.ready();
		}

		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
	}
}