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>
  using min_heap = priority_queue<T, vector<T>, greater<T>>;
  template <typename T>
  using max_heap = priority_queue<T>;
  template <typename T>
  ostream &operator<<(ostream &os, min_heap<T> H) {
    while (!H.empty()) {
      os << H.top() << " ";
      H.pop();
    }
    os << endl;
    return os << "";
  }
  template <typename T>
  ostream &operator<<(ostream &os, max_heap<T> H) {
    while (!H.empty()) {
      os << H.top() << " ";
      H.pop();
    }
    os << endl;
    return os << "";
  }
  template <typename F, typename S>
  ostream &operator<<(ostream &os, pair<F, S> P) {
    return os << P.first << " " << P.second;
  }
  template <typename T>
  ostream &operator<<(ostream &os, vector<T> arr) {
    for (long long i = 0; i < (long long)arr.size(); i++) {
      os << arr[i] << " ";
    }
    return os << "";
  }
  template <typename T>
  ostream &operator<<(ostream &os, vector<vector<T>> matrix) {
    os << endl;
    for (long long i = 0; i < (long long)matrix.size(); i++) {
      for (long long j = 0; j < (long long)matrix[i].size(); j++) {
        os << matrix[i][j] << " ";
      }
      os << endl;
    }
    return os << "";
  }
  template <typename T>
  ostream &operator<<(ostream &os, set<T> S) {
    for (auto s : S) {
      os << s << " ";
    }
    return os << "";
  }
  template <typename T>
  ostream &operator<<(ostream &os, multiset<T> S) {
    for (auto s : S) {
      os << s << " ";
    }
    return os << "";
  }
  template <typename F, typename S>
  ostream &operator<<(ostream &os, map<F, S> M) {
    os << endl;
    for (auto m : M) {
      os << m << endl;
    }
    return os << "";
  }
  template <typename F, typename S>
  ostream &operator<<(ostream &os, multimap<F, S> M) {
    os << endl;
    for (auto m : M) {
      os << m << endl;
    }
    return os << "";
  }
  void solve() {
    long long n, m;
    cin >> n >> m;
    vector<long long> arr(n);
    for (long long i = 0; i < n; i++) {
      cin >> arr[i];
    }
    vector<vector<long long>> pre(n, vector<long long>(m));
    for (long long i = 0; i < (long long)n; i++) {
      string str;
      cin >> str;
      for (long long j = 0; j < (long long)m; j++) {
        pre[i][j] = (str[j] == '1');
      }
    }
    long long mxv = INT64_MIN;
    long long best = -1;
    for (long long mask = 0; mask < (long long)(1 << n); mask++) {
      long long cur = 0;
      for (long long i = 0; i < (long long)n; i++) {
        if ((mask & (1 << i))) {
          cur += arr[i];
        } else {
          cur -= arr[i];
        }
      }
      vector<long long> contri(m);
      for (long long i = 0; i < (long long)n; i++) {
        for (long long j = 0; j < (long long)m; j++) {
          if (pre[i][j]) {
            if (mask & (1 << i)) {
              contri[j]--;
            } else {
              contri[j]++;
            }
          }
        }
      }
      sort((contri).begin(), (contri).end());
      for (long long i = 1; i <= (long long)m; i++) {
        cur += i * contri[i - 1];
      }
      if (mxv < cur) {
        mxv = cur;
        best = mask;
      }
    }
    vector<pair<long long, long long>> contri(m);
    for (long long i = 0; i < (long long)n; i++) {
      for (long long j = 0; j < (long long)m; j++) {
        contri[j].second = j;
        if (pre[i][j]) {
          if (best & (1 << i)) {
            contri[j].first--;
          } else {
            contri[j].first++;
          }
        }
      }
    }
    sort((contri).begin(), (contri).end());
    vector<long long> ans(m);
    for (long long i = 0; i < (long long)m; i++) {
      ans[contri[i].second] = i + 1;
    }
    cout << ans << "\n";
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long no_of_testcases = 1;
    cin >> no_of_testcases;
    for (long long tc = 1; tc <= no_of_testcases; tc++) {
      solve();
    }
    return 0;
  }
  