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 __print(int x) { cout << x; }
  void __print(long x) { cout << x; }
  void __print(long long x) { cout << x; }
  void __print(unsigned x) { cout << x; }
  void __print(unsigned long x) { cout << x; }
  void __print(unsigned long long x) { cout << x; }
  void __print(float x) { cout << x; }
  void __print(double x) { cout << x; }
  void __print(long double x) { cout << x; }
  void __print(char x) { cout << '\'' << x << '\''; }
  void __print(const char *x) { cout << '\"' << x << '\"'; }
  void __print(const string &x) { cout << '\"' << x << '\"'; }
  void __print(bool x) { cout << (x ? "true" : "false"); }
  template <typename T, typename V>
  void __print(const pair<T, V> &x) {
    cout << '{';
    __print(x.first);
    cout << ',';
    __print(x.second);
    cout << '}';
  }
  template <typename T>
  void __print(const T &x) {
    int f = 0;
    cout << '{';
    for (auto &i : x) cout << (f++ ? "," : ""), __print(i);
    cout << "}";
  }
  void _print() { cout << "]\n"; }
  template <typename T, typename... V>
  void _print(T t, V... v) {
    __print(t);
    if (sizeof...(v)) cout << ", ";
    _print(v...);
  }
  const int d4i[4] = {-1, 0, 1, 0}, d4j[4] = {0, 1, 0, -1};
  const int d8i[8] = {-1, -1, 0, 1, 1, 1, 0, -1},
            d8j[8] = {0, 1, 1, 1, 0, -1, -1, -1};
  template <class T>
  bool ckmax(T &a, const T &b) {
    return a < b ? a = b, 1 : 0;
  }
  template <class T>
  bool ckmin(T &a, const T &b) {
    return a > b ? a = b, 1 : 0;
  }
  void solve() {
    int n, m;
    cin >> n >> m;
    vector<int> x(n);
    for (auto &y : x) cin >> y;
    vector<vector<int> > q(n, vector<int>(m));
    for (auto &v : q) {
      string s;
      cin >> s;
      for (int i = 0; i < m; ++i) v[i] = s[i] - '0';
    }
    vector<int> sol(m);
    int best = -1;
    for (int mask = 0; mask < (1 << n); ++mask) {
      int cost = 0;
      for (int i = 0; i < n; ++i) {
        int bit = (1 << i) & mask;
        cost += x[i] * (bit == 0 ? 1 : -1);
      }
      vector<int> qc(m);
      for (int i = 0; i < m; ++i)
        for (int j = 0; j < n; ++j) {
          int bit = (1 << j) & mask;
          if (q[j][i]) qc[i] += bit ? 1 : -1;
        }
      vector<int> ids(m);
      iota((ids).begin(), (ids).end(), 0);
      sort((ids).begin(), (ids).end(),
           [&](int a, int b) { return qc[a] < qc[b]; });
      for (int i = 0; i < m; ++i) cost += (i + 1) * qc[ids[i]];
      if (cost > best) {
        best = cost;
        for (int i = 0; i < m; ++i) sol[ids[i]] = i;
      }
    }
    for (auto y : sol) cout << y + 1 << " ";
    cout << "\n";
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout << fixed << setprecision(12);
    int t;
    cin >> t;
    while (t--) solve();
    return 0;
  }
  