You are given a permutation $p=[p_1, p_2, \ldots, p_n]$ of integers from $1$ to $n$. Let's call the number $m$ ($1 \le m \le n$) beautiful, if there exists two indices $l, r$ ($1 \le l \le r \le n$), such that the numbers $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$.

For example, let $p = [4, 5, 1, 3, 2, 6]$. In this case, the numbers $1, 3, 5, 6$ are beautiful and $2, 4$ are not. It is because:  if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$;  if $l = 3$ and $r = 5$ we will have a permutation $[1, 3, 2]$ for $m = 3$;  if $l = 1$ and $r = 5$ we will have a permutation $[4, 5, 1, 3, 2]$ for $m = 5$;  if $l = 1$ and $r = 6$ we will have a permutation $[4, 5, 1, 3, 2, 6]$ for $m = 6$;  it is impossible to take some $l$ and $r$, such that $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$ for $m = 2$ and for $m = 4$. 

You are given a permutation $p=[p_1, p_2, \ldots, p_n]$. For all $m$ ($1 \le m \le n$) determine if it is a beautiful number or not.


-----Input-----

The first line contains the only integer $t$ ($1 \le t \le 1000$)  — the number of test cases in the input. The next lines contain the description of test cases.

The first line of a test case contains a number $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the given permutation $p$. The next line contains $n$ integers $p_1, p_2, \ldots, p_n$ ($1 \le p_i \le n$, all $p_i$ are different) — the given permutation $p$.

It is guaranteed, that the sum of $n$ from all test cases in the input doesn't exceed $2 \cdot 10^5$.


-----Output-----

Print $t$ lines — the answers to test cases in the order they are given in the input. 

The answer to a test case is the string of length $n$, there the $i$-th character is equal to $1$ if $i$ is a beautiful number and is equal to $0$ if $i$ is not a beautiful number.


-----Example-----
Input
3
6
4 5 1 3 2 6
5
5 3 1 2 4
4
1 4 3 2

Output
101011
11111
1001



-----Note-----

The first test case is described in the problem statement.

In the second test case all numbers from $1$ to $5$ are beautiful:  if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$;  if $l = 3$ and $r = 4$ we will have a permutation $[1, 2]$ for $m = 2$;  if $l = 2$ and $r = 4$ we will have a permutation $[3, 1, 2]$ for $m = 3$;  if $l = 2$ and $r = 5$ we will have a permutation $[3, 1, 2, 4]$ for $m = 4$;  if $l = 1$ and $r = 5$ we will have a permutation $[5, 3, 1, 2, 4]$ for $m = 5$.

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = "t = int(input())\nfor i in range(t):\n    n = int(input())\n    a = [int(i) for i in input().split()]\n    ans = ['0'] * n\n    ans[0] = '1'\n    ans[-1] = '1'\n    l = 0\n    r = n - 1\n    now = n\n    while (r - l) > 1:\n        if a[r] > now:\n            r -= 1\n            continue\n        if a[l] > now:\n            l += 1\n            continue\n        if (r - l + 1) == now:\n            ans[r - l] = '1'\n        now -= 1\n    if (r - l  + 1) == now:\n        ans[r - l] = '1'\n    print(''.join(ans))\n\n\n"
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 13211