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. 

def checker(z,k):
    de=deque()
    for i in range(k):
        if(i==0):
            de.append(i)
        else:
            while(len(de) and z[i]<=z[de[-1]]):
                de.pop()
            de.append(i)

    ans=[]
    for i in range(k,len(z)):
        inter=i-k
        while(de[0]<inter):
            de.popleft()
            
        t=de[0]
        ans.append(z[t])
        while(len(de) and z[i]<=z[de[-1]]):
            de.pop()
        de.append(i)

    while(de[0]<len(z)-k):
        de.popleft()

    t=de[0]
    ans.append(z[t])
    ans.sort()
    for i in range(len(ans)):
        if(ans[i]!=i+1):
            return 0;
    return 1;
            

from collections import *
a=int(input())
for i in range(a):
    s=int(input())
    z=list(map(int,input().split()))
    al=defaultdict(int)
    for i in range(len(z)):
        al[z[i]]+=1
    ans=[0 for i in range(len(z))]

    l=0
    r=s-1
    save=-1
    for i in range(1,s+1):
        if(al[i]==0):
            save=i-1
            break;
        elif(al[i]>1):
            save=i
            break;
        save=i
    compression=len(z)
    for i in range(1,save+1):
        if(z[l]==i and l<=r):
            ans[len(z)-i]=1
            l+=1
            continue;
        elif(z[r]==i and l<=r):
            ans[len(z)-i]=1
            r-=1
            continue;
        else:
            compression=len(z)-i+1
            break;

    z.sort()
    flag=0
 
    for i in range(len(z)):
        if(z[i]!=i+1):
            flag=1
            break;
        
    if(flag==0):
        ans[0]=1

    if(checker(z,compression)==1):
        ans[compression-1]=1
    for i in range(len(ans)):
        print(ans[i],end="")
    print('')
