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: #include <bits/stdc++.h>
  using namespace std;
  long long mod = (1 ? 1e9 + 7 : 998244353);
  const int N = 5e5 + 5;
  void Evlos() {
    int n, m;
    cin >> n >> m;
    vector<int> x(n);
    for (int i = 0; i < n; i++) {
      cin >> x[i];
    }
    vector<vector<int>> a(n, vector<int>(m));
    for (int i = 0; i < n; i++) {
      string s;
      cin >> s;
      for (int j = 0; j < m; j++) {
        a[i][j] = s[j] - '0';
      }
    }
    long long sol = -1;
    vector<int> out(m, -1);
    for (int mask = 0; mask < (1 << n); mask++) {
      vector<int> koef(m);
      long long ans = 0;
      for (int i = 0; i < n; i++) {
        if ((1 << i) & mask) {
          for (int j = 0; j < m; j++) koef[j] += a[i][j];
          ans -= x[i];
        } else {
          for (int j = 0; j < m; j++) koef[j] -= a[i][j];
          ans += x[i];
        }
      }
      vector<int> order(m);
      iota(order.begin(), order.end(), 0);
      sort(order.begin(), order.end(),
           [&](int i, int j) { return koef[i] < koef[j]; });
      vector<int> can(m);
      for (int i = 0; i < m; i++) {
        can[order[i]] = i + 1;
        ans += koef[order[i]] * (i + 1);
      }
      if (ans > sol) {
        sol = ans;
        out = can;
      }
    }
    for (int i = 0; i < m; i++) cout << out[i] << ' ';
    cout << '\n';
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cerr.tie(0);
    int tt = 1;
    cin >> tt;
    while (tt--) Evlos();
    return 0;
  }
  