Consider all binary strings of length $m$ ($1 \le m \le 60$). A binary string is a string that consists of the characters 0 and 1 only. For example, 0110 is a binary string, and 012aba is not. Obviously, there are exactly $2^m$ such strings in total.

The string $s$ is lexicographically smaller than the string $t$ (both have the same length $m$) if in the first position $i$ from the left in which they differ, we have $s[i] < t[i]$. This is exactly the way strings are compared in dictionaries and in most modern programming languages when comparing them in a standard way. For example, the string 01011 is lexicographically smaller than the string 01100, because the first two characters are the same, and the third character in the first string is less than that in the second.

We remove from this set $n$ ($1 \le n \le \min(2^m-1, 100)$) distinct binary strings $a_1, a_2, \ldots, a_n$, each of length $m$. Thus, the set will have $k=2^m-n$ strings. Sort all strings of the resulting set in lexicographical ascending order (as in the dictionary).

We number all the strings after sorting from $0$ to $k-1$. Print the string whose index is $\lfloor \frac{k-1}{2} \rfloor$ (such an element is called median), where $\lfloor x \rfloor$ is the rounding of the number down to the nearest integer.

For example, if $n=3$, $m=3$ and $a=[$010, 111, 001$]$, then after removing the strings $a_i$ and sorting, the result will take the form: $[$000, 011, 100, 101, 110$]$. Thus, the desired median is 100.


-----Input-----

The first line contains an integer $t$ ($1 \le t \le 1000$) — the number of test cases. Then, $t$ test cases follow.

The first line of each test case contains integers $n$ ($1 \le n \le \min(2^m-1, 100)$) and $m$ ($1 \le m \le 60$), where $n$ is the number of strings to remove, and $m$ is the length of binary strings. The next $n$ lines contain $a_1, a_2, \ldots, a_n$ — distinct binary strings of length $m$.

The total length of all given binary strings in all test cases in one test does not exceed $10^5$.


-----Output-----

Print $t$ answers to the test cases. For each test case, print a string of length $m$ — the median of the sorted sequence of remaining strings in the corresponding test case.


-----Example-----
Input
5
3 3
010
001
111
4 3
000
111
100
011
1 1
1
1 1
0
3 2
00
01
10

Output
100
010
0
1
11



-----Note-----

The first test case is explained in the statement.

In the second test case, the result after removing strings and sorting is $[$001, 010, 101, 110$]$. Therefore, the desired median is 010.

```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 = '\nimport bisect\nT = int(input())\n\nfor _ in range(T):\n    N, M = list(map(int, input().split()))\n    aa = sorted([int(input(), 2) for _ in range(N)])\n\n    K = 1 << M\n\n    ans = 0\n\n    for k in reversed(list(range(M))):\n        tmp = ans + (1 << k)\n        La = bisect.bisect_left(aa, tmp)\n        Ra = N - La\n\n        L = tmp - La\n        R = (K - tmp) - Ra\n\n        if L < R:\n            ans = tmp\n        else:\n            continue\n\n    print(f"{ans:0{M}b}")\n'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 13189