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. 
from __future__ import division, print_function

MOD = 998244353
mod = 10**9 + 7

# import resource
# resource.setrlimit(resource.RLIMIT_STACK, [0x100000000, resource.RLIM_INFINITY])

def prepare_factorial():
    fact = [1]
    for i in range(1, 20):
        fact.append((fact[-1] * i) % mod)
    ifact = [0] * 105
    ifact[104] = pow(fact[104], mod - 2, mod)
    for i in range(104, 0, -1):
        ifact[i - 1] = (i * ifact[i]) % mod

    return fact, ifact

# import threading
# threading.stack_size(1<<27)
import sys
# sys.setrecursionlimit(10000)

from bisect import bisect_left, bisect_right, insort
from math import floor, ceil, sqrt, degrees, atan, pi, log, sin, radians, factorial
from heapq import heappop, heapify, heappush
from collections import Counter, defaultdict, deque
from itertools import permutations


def modinv(n, p):
    return pow(n, p - 2, p)

def ncr(n, r, fact, ifact):    # for using this uncomment the lines calculating fact and ifact
    t = (fact[n] * (ifact[r] * ifact[n-r]) % mod) % mod
    return t

def get_ints(): return map(int, sys.stdin.readline().strip().split())
def get_array(): return list(map(int, sys.stdin.readline().strip().split()))
def input(): return sys.stdin.readline().strip()

"""*****************************************************************************************"""

def GCD(x, y):
    while y:
        x, y = y, x % y
    return x

def lcm(x, y):
    return (x * y)//(GCD(x, y))

def get_xor(n):
    return [n, 1, n+1, 0][n % 4]

def get_n(P):    # this function returns the maximum n for which Summation(n) <= Sum
    ans = (-1 + sqrt(1 + 8*P))//2
    return ans

""" ********************************************************************************************* """

def main():

    # def solve(Arr, n):
    #     mat = [[0 for i in range(n)] for j in range(n)]
    #     for i in range(n):
    #         mat[0][i] = Arr[i]
    #
    #     for i in range(1, n):
    #         for j in range(0, n-i):
    #             mat[i][j] = min(mat[i-1][j], mat[i-1][j+1])
    #
    #     # for i in mat:
    #     #     print(i)
    #
    #     ans = ['0']*n
    #     for i in range(n):
    #         myset = set()
    #         for j in range(0, n-i):
    #             myset.add(mat[i][j])
    #         if len(myset) == n-i:
    #             ans[i] = '1'
    #     if ans[-4] == '1':
    #         print(Arr, "-----", ''.join(ans))


    T = int(input())
    while T:
        n = int(input())
        Arr = get_array()

        ans = ['0']*n
        mydict1 = defaultdict(int)
        mydict2 = defaultdict(list)

        for index, value in enumerate(Arr):
            mydict1[value] += 1
            mydict2[value].append(index)
        left = 0
        right = n-1
        for i in range(1, n):
            if mydict1[i] == 0:
                break
            elif mydict1[i] == 1:
                ans[-i] = '1'
                if mydict2[i][0] == left:
                    left += 1
                elif mydict2[i][0] == right:
                    right -= 1
                else:
                    break
            else:
                ans[-i] = '1'
                break

        if len(set(Arr)) == n:
            ans[-n] = '1'
        print(''.join(ans))
        T -= 1




""" -------- Python 2 and 3 footer by Pajenegod and c1729 ---------"""

py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange

import os, sys
from io import IOBase, BytesIO

BUFSIZE = 8192

class FastIO(BytesIO):
    newlines = 0

    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None

    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0, 2), super(FastIO, self).write(s))[0])
        return s

    def read(self):
        while self._fill(): pass
        return super(FastIO, self).read()

    def readline(self):
        while self.newlines == 0:
            s = self._fill();
            self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)


class IOWrapper(IOBase):

    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            self.write = lambda s: self.buffer.write(s.encode('ascii'))
            self.read = lambda: self.buffer.read().decode('ascii')
            self.readline = lambda: self.buffer.readline().decode('ascii')


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')

# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')

if __name__ == '__main__':
    main()
    # threading.Thread(target=main).start()