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;
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  template <typename T>
  istream &operator>>(istream &in, vector<T> &v) {
    for (auto &x : v) in >> x;
    return in;
  }
  template <typename T>
  ostream &operator<<(ostream &out, const vector<T> &v) {
    for (auto &x : v) out << x << ' ';
    return out;
  }
  template <typename T1, typename T2>
  istream &operator>>(istream &in, pair<T1, T2> &p) {
    in >> p.first >> p.second;
    return in;
  }
  template <typename T1, typename T2>
  ostream &operator<<(ostream &out, const pair<T1, T2> &p) {
    out << p.first << ' ' << p.second;
    return out;
  }
  const bool tests = 1;
  void solve_case() {
    long long n, m;
    cin >> n >> m;
    vector<long long> x(n);
    cin >> x;
    vector<string> s(n);
    cin >> s;
    vector<long long> ans(m);
    long long ansv = -1;
    for (int mask = 0; mask < (1 << n); mask++) {
      vector<long long> cnt(m, 0);
      for (int j = 0; j < n; j++) {
        for (int k = 0; k < m; k++) {
          if (s[j][k] == '1') cnt[k] += 2 * ((mask & (1 << j)) > 0) - 1;
        }
      }
      vector<long long> vec(m);
      iota(vec.begin(), vec.end(), 0);
      sort(vec.begin(), vec.end(), [&](const long long &a, const long long &b) {
        return cnt[a] < cnt[b];
      });
      vector<long long> q(m);
      for (int i = 0; i < m; i++) {
        q[vec[i]] = i;
      }
      bool flag = 0;
      long long tempans = 0;
      for (int j = 0; j < n; j++) {
        long long sm = 0;
        for (int k = 0; k < m; k++) {
          if (s[j][k] == '1') sm += q[k] + 1;
        }
        if (mask & (1 << j)) {
          if (sm < x[j]) {
            flag = 1;
            break;
          }
          tempans += sm - x[j];
        } else {
          if (sm > x[j]) {
            flag = 1;
            break;
          }
          tempans += x[j] - sm;
        }
      }
      if (flag) continue;
      if (ansv < tempans) {
        ansv = tempans;
        ans = q;
      }
    }
    for (long long x : ans) cout << ++x << ' ';
    cout << "\n";
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long testcase = 1;
    if (tests) cin >> testcase;
    cout << fixed << setprecision(10);
    for (long long test = 1; test <= testcase; test++) {
      solve_case();
    }
    return 0;
  }
  