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 sys
#from collections import deque
#from functools import lru_cache
#from fractions import Fraction as f
from copy import *
from bisect import *	
#from heapq import *
from math import gcd,ceil,sqrt
#from itertools import permutations as prm,product
#from random import *
def eprint(*args):
    print(*args, file=sys.stderr)
zz=1
 
sys.setrecursionlimit(10**5)
if zz:
	input=sys.stdin.readline
else:	
	sys.stdin=open('input.txt', 'r')
	sys.stdout=open('all.txt','w')
di=[[-1,0],[1,0],[0,1],[0,-1]]

def string(s):
	return "".join(s)
def fori(n):
	return [fi() for i in range(n)]	
def inc(d,c,x=1):
	d[c]=d[c]+x if c in d else x
def li():
	return [int(xx) for xx in input().split()]
def fli():
	return [float(x) for x in input().split()]	
def comp(a,b):
	if(a>b):
		return 2
	return 2 if a==b else 0		
def gi():	
	return [xx for xx in input().split()]
def gtc(tc,ans):
	print("Case #"+str(tc)+":",ans)	
def cil(n,m):
	return n//m+int(n%m>0)	
def fi():
	return int(input())
def pro(a): 
	return reduce(lambda a,b:a*b,a)		
def swap(a,i,j): 
	a[i],a[j]=a[j],a[i]	
def si():
	return list(input().rstrip())	
def mi():
	return 	map(int,input().split())			
def gh():
	sys.stdout.flush()
def isvalid(i,j,n,m):
	return 0<=i<n and 0<=j<m 
def bo(i):
	return ord(i)-ord('a')	
def graph(n,m):
	for i in range(m):
		x,y=mi()
		a[x].append(y)
		a[y].append(x)


t=fi()

		
while t>0:
	t-=1
	n=fi()
	a=li()
	st=[]
	pre=[n]*n
	for i in range(n):
		if len(st)==0:
			st.append(i)
		else:
			if a[st[-1]]>a[i]:
				while len(st) and a[st[-1]]>a[i]:
					pre[st[-1]]=i
					st.pop()
			st.append(i)
	#print(pre)		
	st=[]
	suf=[-1]*n	
	for i in range(n-1,-1,-1):
		if len(st)==0:
			st.append(i)
		else:
			if a[st[-1]]>a[i]:
				while len(st) and a[st[-1]]>a[i]:
					suf[st[-1]]=i
					st.pop()
			st.append(i)
	#print(suf)
	d=[0]*(n+1)
	for i in range(n):
		d[a[i]]=max(d[a[i]],pre[i]-suf[i]-1)
	ans=""
	flag=[1]*(n+1)
	pm=[n for i in range(n+1)]
	for i in range(1,n+1):
		pm[i]=min(pm[i-1],d[i])
	for i in range(1,n+1):
		if pm[n-i+1]<i:
			flag[i]=0
	for i in range(1,n+1):
		print(flag[i],end="")
	print()		

