human
Consider a closed world and a set of features that are defined for all the objects in the world. Each feature can be answered with "yes" or "no". Using those features, we can identify any object from the rest of the objects in the world. In other words, each object can be represented as a fixed-length sequence of booleans. Any object is different from other objects by at least one feature.

You would like to identify an object from others. For this purpose, you can ask a series of questions to someone who knows what the object is. Every question you can ask is about one of the features. He/she immediately answers each question with "yes" or "no" correctly. You can choose the next question after you get the answer to the previous question.

You kindly pay the answerer 100 yen as a tip for each question. Because you don' t have surplus money, it is necessary to minimize the number of questions in the worst case. You don’t know what is the correct answer, but fortunately know all the objects in the world. Therefore, you can plan an optimal strategy before you start questioning.

The problem you have to solve is: given a set of boolean-encoded objects, minimize the maximum number of questions by which every object in the set is identifiable.


Input

The input is a sequence of multiple datasets. Each dataset begins with a line which consists of two integers, m and n: the number of features, and the number of objects, respectively. You can assume 0 < m ≤ 11 and 0 < n ≤ 128. It is followed by n lines, each of which corresponds to an object. Each line includes a binary string of length m which represent the value ("yes" or "no") of features. There are no two identical objects.

The end of the input is indicated by a line containing two zeros. There are at most 100 datasets.

Output

For each dataset, minimize the maximum number of questions by which every object is identifiable and output the result.

Example

Input

8 1
11010101
11 4
00111001100
01001101011
01010000011
01100110001
11...
impor
=====
Here the object is?
assistant
#Here the object is.
t math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools

sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**13
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]

def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)

def main():
    rr = []

    def f(n,m):
        a = [int(S(), 2) for _ in range(m)]
        mi = 0
        ma = max(a)
        while 2**mi <= ma:
            mi += 1
        ii = [2**i for i in range(mi+1)]

        fm = {}
        def _f(a):
            k = tuple(a)
            if k in fm:
                return fm[k]

            if len(a) < 2:
                fm[k] = 0
                return 0

            r = inf
            for i in ii:
                a1 = []
                a2 = []
                for c in a:
                    if i & c:
                        a1.append(c)
                    else:
                        a2.append(c)
                if not a1 or not a2:
                    continue
                r1 = _f(a1)
                r2 = _f(a2)
                tr = max(r1, r2) + 1
                if r > tr:
                    r = tr
            fm[k] = r
            return r

        r = _f(a)
        return r

    while 1:
        n,m = LI()
        if n == 0:
            break
        rr.append(f(n,m))

    return '\n'.join(map(str,rr))

print(main())