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,math
input = sys.stdin.readline
tree = [0]
def rin():
    return int(sys.stdin.readline().strip())

  

  
def build(arr) :
  
    # insert leaf nodes in tree 
    for i in range(n) : 
        tree[n + i] = arr[i]; 
      
    # build the tree by calculating parents 
    for i in range(n - 1, 0, -1) : 
        tree[i] = min(tree[i << 1], tree[i << 1 | 1])

# function to get sum on interval [l, r) 
def query(l, r) : 
  
    res = 1000000; 
      
    # loop to find the sum in the range 
    l += n;
    r += n;
      
    while l < r :
      
        if (l & 1) :
            res = min(res,tree[l])
            l += 1
      
        if (r & 1) :
            r -= 1
            res = min(res,tree[r]) 
              
        l >>= 1
        r >>= 1
      
    return res 
def solution(A):
    return min(A) == 1 and max(A) == len(A) == len(set(A))

t = rin()
for _ in range(t):
    n = rin()
    arr = [int(i) for i in input().strip().split()]
    tree = [0] * (math.ceil(math.log2(n))*n)
    tree.append(0)
    tree.append(0)
    build(arr)
    ans = []
    l = 0
    r = n - 1
    for i in range(1, n):
        t_min = query(l, r + 1)
        #print(i,'left: ', l, 'right: ',r, 'mini :',t_min)
        if t_min != i:
            for i in range(i, n):
                ans.append('0')
            break
        if arr[l] == i:
            l += 1
            ans.append('1')
        elif arr[r] == i:
            r -= 1
            ans.append('1')
        else:
            ans.append('1')
            for i in range(i + 1, n):
                ans.append('0')
            break
    if solution(arr):
        ans.append('1')
    else:
        ans.append('0')
    ans.reverse()
    print(''.join(ans))
