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;
  template <typename T, typename S>
  ostream& operator<<(ostream& out, const pair<T, S>& p) {
    out << "(" << p.first << " " << p.second << ")";
    return out;
  }
  template <typename T>
  ostream& operator<<(ostream& out, const vector<T>& v) {
    out << "[";
    for (int i = 0; i < (int)v.size(); i++) {
      out << v[i];
      if (i != (int)v.size() - 1) out << ", ";
    }
    out << "]";
    return out;
  }
  template <typename T>
  ostream& operator<<(ostream& out, const vector<vector<T> >& m) {
    for (int i = 0; i < m.size(); ++i) {
      out << m[i] << "\n";
    }
    return out;
  }
  template <typename H1>
  ostream& show(ostream& out, const char* label, H1&& value) {
    return out << label << " = " << value << endl;
  }
  template <typename H1, typename... T>
  ostream& show(ostream& out, const char* label, H1&& value, T&&... rest) {
    const char* first_comma = strchr(label, ',');
    const char* left_parenthesis = strchr(label, '(');
    if (left_parenthesis != nullptr && left_parenthesis < first_comma) {
      const char* right_parenthesis = strchr(left_parenthesis, ')');
      assert(right_parenthesis != nullptr);
      const char* pcomma = strchr(right_parenthesis, ',');
      return show(out.write(label, pcomma - label) << " = " << value << ',',
                  pcomma + 1, rest...);
    }
    return show(out.write(label, first_comma - label) << " = " << value << ',',
                first_comma + 1, rest...);
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int cases;
    cin >> cases;
    for (int ic = 0; ic < cases; ++ic) {
      int n, m;
      cin >> n >> m;
      vector<int> expect(n);
      for (int& x : expect) {
        cin >> x;
      }
      vector<string> result(n);
      for (string& x : result) {
        cin >> x;
      }
      int ans = -1;
      vector<pair<int, int> > permutation;
      for (int i = 0; i < (1 << n); ++i) {
        int tmp_ans = 0;
        vector<pair<int, int> > weight(m);
        for (int j = 0; j < m; ++j) {
          weight[j] = {0, j};
        }
        for (int j = 0; j < n; ++j) {
          int sign = (i & (1 << j) ? 1 : -1);
          for (int k = 0; k < m; ++k) {
            weight[k].first += (result[j][k] - '0') * sign;
          }
          tmp_ans += -sign * expect[j];
        }
        sort(weight.begin(), weight.end());
        for (int j = 1; j <= m; ++j) {
          tmp_ans += j * weight[j - 1].first;
        }
        if (ans < tmp_ans) {
          ans = tmp_ans;
          swap(permutation, weight);
        }
      }
      vector<int> per_ans(m);
      for (int i = 0; i < m; ++i) {
        per_ans[permutation[i].second] = i + 1;
      }
      for (int x : per_ans) {
        cout << x << ' ';
      }
      cout << '\n';
    }
  }
  