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
input=sys.stdin.buffer.readline #FOR READING PURE INTEGER INPUTS (space separation ok)

import math 
#MIN QUERIES
class RMQMIN(): #for MIN queries
    def __init__(self,arr):
        self.arr=arr
        MAXN=len(arr)
        K=int(math.log2(MAXN))+1
        self.lookup=[[0 for _ in range(K+1)] for __ in range(MAXN)]
        self.buildSparseTable()
    def buildSparseTable(self): 
        n=len(self.arr)
        for i in range(0, n):  
            self.lookup[i][0] = self.arr[i]  
        j = 1
        while (1 << j) <= n:
            i = 0
            while (i + (1 << j) - 1) < n:
                if (self.lookup[i][j-1]<self.lookup[i+(1<<(j-1))][j-1]):self.lookup[i][j]=self.lookup[i][j-1]  
                else:self.lookup[i][j]=self.lookup[i + (1 << (j - 1))][j - 1]
                i += 1
            j += 1
    def query(self,L,R): #returns min on interval [l,r]
        j = int(math.log2(R - L + 1))
        if self.lookup[L][j] <= self.lookup[R - (1 << j) + 1][j]:  
            return self.lookup[L][j]
        else: 
            return self.lookup[R - (1 << j) + 1][j]  
#min query usage:
#a = [7, 2, 3, 0, 5,1,3,4,63,2,4,5,6,7,22]  
#rmq=RMQMIN(a)
#rmq.query(6,len(a))

t=int(input())
for _ in range(t):
    n=int(input())
    a=[int(x) for x in input().split()]
    
    ans=[0 for _ in range(n+1)]
    
    #check for k=1
    if sorted(a)==list(range(1,n+1)):
        ans[1]=1
    
    rmq=RMQMIN(a)
    l=0
    r=n-1
    #for k==n, 1 has to be in the array
    #for k==n-1 to make CodeCook people happy, 1 has to be at either end,rmq of the rest is 2
    #for k==n-2, 2 has to be at either end, rmq of the rest is 3, and so on...
    if rmq.query(l,r)==1:
        ans[n]=1
    k=n-1
    smallest=1
    while k>1:
        if a[l]==smallest and rmq.query(l+1,r)==smallest+1:
            l+=1
            smallest+=1
            ans[k]=1
            k-=1
        elif a[r]==smallest and rmq.query(l,r-1)==smallest+1:
            r-=1
            smallest+=1
            ans[k]=1
            k-=1
        else:
            break
    ans2=[]
    for i in range(1,n+1):
        ans2.append(str(ans[i]))
    print(''.join(ans2))