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 <class a, class b>
  ostream& operator<<(ostream& os, const pair<a, b>& p) {
    os << "(" << p.first << "," << p.second << ")";
    return os;
  }
  template <class T>
  ostream& operator<<(ostream& os, const vector<T>& v) {
    os << "[";
    if (!v.size())
      os << "]";
    else
      for (int i = 0; i < v.size(); ++i) os << v[i] << ",]"[i == v.size() - 1];
    return os;
  }
  template <class T>
  ostream& operator<<(ostream& os, const set<T>& s) {
    os << "{";
    if (!s.size())
      os << "}";
    else
      for (auto x : s) os << x << ",}"[x == *s.rbegin()];
    return os;
  }
  void printr(ostream& os) { os << '\n'; }
  template <class T, class... Args>
  void printr(ostream& os, const T& t, const Args&... args) {
    os << t;
    if (sizeof...(args)) os << " ";
    printr(os, args...);
  }
  template <class T>
  void read(T& x) {
    cin >> x;
  }
  template <class T, class... Args>
  void read(T& t, Args&... args) {
    read(t), read(args...);
  }
  template <class T>
  void readarray(T* A, int n) {
    for (__typeof(0) i = (0), _end = (n), _step = 1 - 2 * ((0) > (n)); i != _end;
         i += _step)
      read(A[i]);
  }
  void dbgr(ostream& os) { os << endl; }
  template <class T, class... Args>
  void dbgr(ostream& os, const T& t, const Args&... args) {
    os << t;
    if (sizeof...(args)) os << ", ";
    dbgr(os, args...);
  }
  template <class T>
  T min(const vector<T>& v) {
    return *min_element(v.begin(), v.end());
  }
  template <class T>
  T max(const vector<T>& v) {
    return *max_element(v.begin(), v.end());
  }
  long long pw(long long a, long long k = 998244353 - 2) {
    long long ans = 1;
    for (a %= 998244353; k; k >>= 1) {
      if (k & 1) ans = ans * a % 998244353;
      a = a * a % 998244353;
    }
    return ans;
  }
  int n, m;
  int x[15];
  char a[15][10050];
  int pos[10050], p[10050];
  long long c[10050];
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int _;
    read(_);
    while (_--) {
      cin >> n >> m;
      for (int i = 0; i < n; ++i) cin >> x[i];
      for (int i = 0; i < n; ++i) cin >> a[i];
      long long ans = -1e18;
      for (int s = 0; s < (1 << n); ++s) {
        long long sum = 0;
        for (int j = 0; j < m; ++j) c[j] = 0, pos[j] = j;
        for (int i = 0; i < n; ++i) {
          long long sgn = (s >> i & 1) ? 1 : -1;
          sum -= x[i] * sgn;
          for (int j = 0; j < m; ++j)
            if (a[i][j] == '1') c[j] += sgn;
        }
        sort(pos, pos + m, [&](int i, int j) { return c[i] < c[j]; });
        for (int j = 0; j < m; ++j) sum += (j + 1) * c[pos[j]];
        if (sum > ans) {
          ans = sum;
          for (int j = 0; j < m; ++j) p[pos[j]] = j + 1;
        }
      }
      for (int j = 0; j < m; ++j) cout << p[j] << " ";
      cout << endl;
    }
    return 0;
  }
  