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.readline
def fun(l,r,val,che,ar):
    while(l<=r):
        mid=l+(r-l)//2
        if(fun1(mid,ar)<val):
            l=mid+1
        else:
            if(che[mid]==1):
                return mid
            else:
                r=mid-1
def update(inp,add,ar):
    global n
    while(inp<n):
        ar[inp]+=add
        inp+=(inp&(-inp))
def fun1(inp,ar):
    global n
    ans=0
    while(inp):
        ans+=ar[inp]
        inp-=(inp&(-inp))
    return ans
for _ in range(int(input())):
    nn=int(input())
    n=nn+1
    br=list(map(int,input().split()))
    ar=[0]*n
    arb=[0]*n
    ch=[0]*n
    chb=[0]*n
    dic={}
    ans=[-1]*n
    for i in range(nn):
        if(br[i] in dic):
            dic[br[i]].append(i)
        else:
            dic[br[i]]=[i]
    for i in range(1,n):
        if(i in dic):
            for j in dic[i]:
                temval=fun1(j+1,ar)
                if(temval==0):
                    fl=0
                else:
                    fl=fun(0,j+1,fun1(j+1,ar),ch,ar)
                j=(nn-1-j)
                temval=fun1(j+1,arb)
                if(temval==0):
                    rl=0
                else:
                    rl=fun(0,j+1,fun1(j+1,arb),chb,arb)
                ans[i]=max(ans[i],nn-(rl+fl))
            for j in dic[i]:
                ch[j+1]=1
                chb[nn-j]=1
                update(j+1,1,ar)
                update(nn-j,1,arb)
    mi=[0]
    va=ans[1]
    for i in range(1,n):
        va=min(va,ans[i])
        mi.append(va)
    final=[0]*n
    for i in range(1,n):
        req=n-i
        if(mi[i]>=req):
            final[req]=1
    for i in range(1,n):
        print(final[i],end='')
    print()