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;
  void read(vector<int64_t> &a);
  void read(vector<string> &a);
  void read(vector<vector<int64_t>> &a);
  void read(vector<pair<int64_t, int64_t>> &a);
  void solve() {
    int64_t n, m;
    cin >> n >> m;
    vector<int64_t> exp(n);
    read(exp);
    vector<string> mat(n);
    read(mat);
    int64_t best = 0;
    vector<int64_t> ret;
    for (int64_t x = 0; x < (1 << n); x++) {
      string b = std::bitset<32>(x).to_string();
      reverse(b.begin(), b.end());
      vector<pair<int64_t, int64_t>> order(m);
      for (int64_t i = 0; i < m; i++) {
        int64_t temp = 0;
        for (int64_t j = 0; j < n; j++) {
          if (b[j] == '1')
            temp += (mat[j][i] - '0');
          else
            temp -= (mat[j][i] - '0');
        }
        order[i] = {temp, i};
      }
      sort(order.begin(), order.end());
      reverse(order.begin(), order.end());
      int64_t ans = 0;
      for (int64_t i = 0; i < n; i++) {
        if (b[i] == '0')
          ans += exp[i];
        else
          ans -= exp[i];
      }
      int64_t tog = m;
      vector<int64_t> haha(m);
      for (auto x : order) {
        ans += (x.first) * (tog);
        haha[x.second] = tog;
        tog--;
      }
      if (ans >= best) {
        best = ans;
        ret = haha;
      }
    }
    for (int64_t i : ret) cout << i << " ";
    cout << endl;
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int64_t t = 1;
    cin >> t;
    for (int64_t tt = 1; tt <= t; tt++) {
      solve();
    }
    return 0;
  }
  void read(vector<int64_t> &a) {
    for (int64_t i = 0; i < a.size(); i++) cin >> a[i];
  }
  void read(vector<string> &a) {
    for (int64_t i = 0; i < a.size(); i++) cin >> a[i];
  }
  void read(vector<vector<int64_t>> &a) {
    for (int64_t i = 0; i < a.size(); i++)
      for (int64_t j = 0; j < a[i].size(); j++) cin >> a[i][j];
  }
  void read(vector<pair<int64_t, int64_t>> &a) {
    for (int64_t i = 0; i < a.size(); i++) cin >> a[i].first >> a[i].second;
  }
  