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
if sys.subversion[0] == "PyPy":
    import io, atexit
    sys.stdout = io.BytesIO()
    atexit.register(lambda: sys.__stdout__.write(sys.stdout.getvalue()))
    
    sys.stdin = io.BytesIO(sys.stdin.read())
    input = lambda: sys.stdin.readline().rstrip()

RS = raw_input
RI = lambda x=int: map(x,RS().split())
RN = lambda x=int: x(RS())
''' ...................................................................... '''
class SegTree:
    ''' PU-RQ '''
 
    def __init__(self,arr,n):
        ''' Builds in O(N) '''
        
        self.N = n
        self.tree = [float('inf')]*(2*n)
 
        for i in xrange(n):     # Leaf Node
            self.tree[n+i] = arr[i]
 
        for i in xrange(n-1,0,-1):  # Parent Node
            self.tree[i] = min(self.tree[2*i] , self.tree[2*i+1])
 
    
    def update(self, i, newValue):
        ''' 0-Based Indexing '''
        
        i += self.N     # Points to leaf node (i)
 
        self.tree[i] = newValue
 
        while i>1:
            i/=2        # Points to parent node
            self.tree[i] = min(self.tree[2*i] , self.tree[2*i+1])
 
 
    def query(self, L, R):
        ''' Query in interval [L,R] (0-Based Indexing) '''
        
        # Leaf Nodes
        L += self.N
        R += self.N + 1
 
        res = float('inf')
        while L<R:
            if L&1:
                res = min(res,self.tree[L])
                #res += self.tree[L]     # Right Child
                L += 1
                
            if R&1:
                R -= 1
                res = min(res,self.tree[R])
                #res += self.tree[R]     # Left Child
                
            L/=2; R/=2
        return res

    
for _ in xrange(RN()):
    n = RN()
    arr = RI()
    seg = SegTree(arr,n)
    
    ans = [0]*n
    lo,hi = 0,n-1

    val = 1
    
    for k in xrange(n,1,-1):
        m = seg.query(lo,hi)

        if m==val:
            ans[k-1]=1

            if arr[lo]==m: lo+=1
            elif arr[hi]==m: hi-=1
            else: break
        else:
            break

        val+=1


    if len(set(arr))==n: ans[0]=1
        
    print ''.join(map(str,ans))
    
