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: import sys
  import time
  import random
  I=lambda:[*map(int,sys.stdin.readline().split())]
  
  t, = I()
  for _ in range(t):
  	n, m = I()
  	expect = I()
  	mat = []
  	for i in range(n):
  		mat.append([*map(int,[*sys.stdin.readline().strip()])])
  	# n, m = 10, 10000
  	# expect = [random.randint(1, 50000000) for i in range(n)]
  	# mat = [[0] * m for i in range(n)]
  	# for i in range(n):
  	# 	for j in range(m):
  	# 		mat[i][j] = random.randint(0, 1)
  
  
  	#print(time.time())
  
  	onebits = []
  	for i in range(1 << n):
  		onebits.append(bin(i).count('1'))
  	qrep = [0] * m
  	for i in range(n):
  		for j in range(m):
  			if mat[i][j]:
  				qrep[j] += 1 << i
  	big = -1
  	perm = None
  	for i in range(1 << n):
  		count = [0] * (n + 1)
  		vals = [0] * m
  		for j in range(m):
  			x = onebits[i ^ qrep[j]]
  			vals[j] = x
  			count[x] += 1
  
  		ind = [0] * (n + 1)
  		for j in range(n):
  			ind[j + 1] = ind[j] + count[j]
  
  		qvals = [0] * m
  		for j in range(m):
  			ind[vals[j]] += 1
  			qvals[j] = ind[vals[j]]
  
  		tot = 0
  		for j in range(n):
  			if (i >> j) & 1:
  				tot += expect[j]
  			else:
  				tot -= expect[j]
  
  		for j in range(m):
  			right = onebits[qrep[j]]
  			x = onebits[i & qrep[j]]
  			tot += (right - 2 * x) * qvals[j]
  		# scores = []
  		# for k in range(n):
  		# 	scores.append(sum(qvals[j] for j in range(m) if mat[k][j]))
  		# surprise = sum(abs(scores[k] - expect[k]) for k in range(n))
  		#print(pos, qvals, surprise)
  		if tot > big:
  			big = tot
  			perm = qvals
  	print(*perm)
  
  #print(time.time())