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

sys.setrecursionlimit(10**5)
int1 = lambda x: int(x)-1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.buffer.readline())
def MI(): return map(int, sys.stdin.buffer.readline().split())
def LI(): return list(map(int, sys.stdin.buffer.readline().split()))
def LI1(): return list(map(int1, sys.stdin.buffer.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def BI(): return sys.stdin.buffer.readline().rstrip()
def SI(): return sys.stdin.buffer.readline().rstrip().decode()
inf = 10**16
md = 10**9+7
# md = 998244353

class SparseTableMin:
    def __init__(self, aa):
        w = len(aa)
        h = w.bit_length()
        table = [aa]+[[-1]*w for _ in range(h-1)]
        tablei1 = table[0]
        for i in range(1, h):
            tablei = table[i]
            for j in range(w-(1 << i)+1):
                rj = j+(1 << (i-1))
                tablei[j] = min(tablei1[j], tablei1[rj])
            tablei1 = tablei
        self.table = table

    # [l,r)の最小値
    def min(self, l, r):
        i = (r-l).bit_length()-1
        tablei = self.table[i]
        Lmin = tablei[l]
        Rmin = tablei[r-(1 << i)]
        if Lmin < Rmin: Rmin = Lmin
        return Rmin

for _ in range(II()):
    n = II()
    aa = LI1()

    ans = [0]*n
    ex = [False]*n
    for a in aa:
        if a < n: ex[a] = True
    if ex[0]: ans[n-1] = 1
    if sum(ex) == n: ans[0] = 1

    sp = SparseTableMin(aa)
    l, r = 0, n-1
    for i in range(n-2):
        if aa[l] == i and sp.min(l+1, r+1) == i+1:
            ans[n-2-i] = 1
            l += 1
        elif aa[r] == i and sp.min(l, r) == i+1:
            ans[n-2-i] = 1
            r -= 1
        else:
            break

    print(*ans,sep="")
