description: Petya is a math teacher. n of his students has written a test consisting of m questions. For each student, it is known which questions he has answered correctly and which he has not.
  
  If the student answers the j-th question correctly, he gets p_j points (otherwise, he gets 0 points). Moreover, the points for the questions are distributed in such a way that the array p is a permutation of numbers from 1 to m.
  
  For the i-th student, Petya knows that he expects to get x_i points for the test. Petya wonders how unexpected the results could be. Petya believes that the surprise value of the results for students is equal to ∑_{i=1}^{n} |x_i - r_i|, where r_i is the number of points that the i-th student has got for the test.
  
  Your task is to help Petya find such a permutation p for which the surprise value of the results is maximum possible. If there are multiple answers, print any of them.
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
  
  The first line of each test case contains two integers n and m (1 ≤ n ≤ 10; 1 ≤ m ≤ 10^4) — the number of students and the number of questions, respectively.
  
  The second line contains n integers x_1, x_2, ..., x_n (0 ≤ x_i ≤ (m(m+1))/(2)), where x_i is the number of points that the i-th student expects to get.
  
  This is followed by n lines, the i-th line contains the string s_i (|s_i| = m; s_{i, j} ∈ \{0, 1\}), where s_{i, j} is 1 if the i-th student has answered the j-th question correctly, and 0 otherwise.
  
  The sum of m for all test cases does not exceed 10^4.
  
  Output
  
  For each test case, print m integers — a permutation p for which the surprise value of the results is maximum possible. If there are multiple answers, print any of them.
  
  Example
  
  Input
  
  
  3
  4 3
  5 1 2 2
  110
  100
  101
  100
  4 4
  6 2 0 10
  1001
  0010
  0110
  0101
  3 6
  20 3 15
  010110
  000101
  111111
  
  
  Output
  
  
  3 1 2 
  2 3 4 1 
  3 1 4 5 2 6 
solution: ''' E. Math Test
  https://codeforces.com/contest/1622/problem/E
  '''
  
  import io, os, sys
  input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline  # decode().strip() if str
  output = sys.stdout.write
  
  DEBUG = os.environ.get('debug') is not None
  
  if DEBUG:
      from inspect import currentframe, getframeinfo
      from re import search
  
  def debug(*args):
      if not DEBUG: return
      frame = currentframe().f_back
      s = getframeinfo(frame).code_context[0]
      r = search(r"\((.*)\)", s).group(1)
      vnames = r.split(', ')
      var_and_vals = [f'{var}={val}' for var, val in zip(vnames, args)]
      prefix = f'{currentframe().f_back.f_lineno:02d}: '
      print(f'{prefix}{", ".join(var_and_vals)}')
  
  
  INF = float('inf')
  
  # -----------------------------------------
  
  def bucket_sort(vals, order=None):
      '''
      given array [vals[i] for i in order]
      return new_order s.t. [vals[i] for i in new_order] is sorted
      '''
      N = len(vals)
      if order is None: order = list(range(N))
  
      # normalize so mn == 0
      mn = min(vals)
      vals = [v-mn for v in vals]
      mx = max(vals)
  
      # count[v] = number of value v in vals
      count = [0]*(mx+1)
      for v in vals: count[v] += 1
  
      # count[v] = number of values in vals <= v
      for i in range(len(count)-1): count[i+1] += count[i]
  
      # new order
      new_order = [0]*N
      for i in reversed(order):
          v = vals[i]
          c = count[v]        # c = num values in vals <= v
          new_order[c-1] = i  # so put v at idx c-1 (0-based)
          count[v] -= 1       # next time encountering v, put it at idx c-2
  
      return new_order
  
  
  # let sign[i] = sign(r[i] - x[i])
  # then total surprise = SUM_i sign[i] * (r[i] - x[i])
  # expectation i contributes: - sign[i] * x[i] ---> fixed
  # question q contributes:      score[q] * SUM_i signed[i] * (whether student i solves question q)
  # -> can greedily assign highest scores to highest contribution questions
  
  # if after assignment, student i has sign(r[i] - x[i]) != sign[i]
  # i.e. subtract |r[i] - x[i]| from total surprise instead of add
  # then flipping sign[i] will increase total surprise
  # so the orig mask cannot be optimal
  
  def solve(N, M, X, A):
      mx, mxv = -INF, []
      for mask in range(1 << N):
          x = 0
  
          # add[q] = num students with mask 1 who solved q - num students with mask 0 who solved q
          add = [0]*M
          for i in range(N):
              sign = 1 if (mask >> i) & 1 == 1 else -1
              x -= sign * X[i]
              for q in range(M):
                  add[q] += A[i][q] * sign
  
          # greedily assign highest score to q with highest add
          # normal sort will TLE
          order = bucket_sort(add)
          x += sum((v+1) * add[q] for v, q in enumerate(order))
          if x > mx: mx, mxv = x, order
  
      res = [-1]*M
      for v, q in enumerate(mxv): res[q] = v+1
      return res
  
  
  def main():
      T = int(input())
      for _ in range(T):
          N, M = list(map(int, input().split()))
          X = list(map(int, input().split()))
          A = [list(map(int, list(input().decode().strip()))) for _ in range(N)]
          out = solve(N, M, X, A)
          print(*out)
  
  
  if __name__ == '__main__':
      main()
   