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;
  using ll = long long;
  using pii = pair<int, int>;
  using pll = pair<ll, ll>;
  using vi = vector<int>;
  using vl = vector<ll>;
  using vii = vector<pii>;
  using vll = vector<pll>;
  void debug_out() { cerr << endl; }
  template <typename Head, typename... Tail>
  void debug_out(Head H, Tail... T) {
    cerr << " " << to_string(H);
    debug_out(T...);
  }
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  const long double eps = 1e-9;
  const long long mod = 998244353;
  const int MAXN = 200000;
  void solve() {
    ll n, m;
    cin >> n >> m;
    pair<ll, string> arr[n];
    for (int i = 0, nnnn = (n); i < nnnn; ++i) cin >> arr[i].first;
    for (int i = 0, nnnn = (n); i < nnnn; ++i) cin >> arr[i].second;
    ll ans = 0;
    ll p[m];
    for (int i = 0, nnnn = (m); i < nnnn; ++i) p[i] = i + 1;
    for (int i = 0, nnnn = ((1 << n)); i < nnnn; ++i) {
      pair<ll, ll> cnt[m];
      memset(cnt, 0, sizeof cnt);
      for (int j = 0, nnnn = (m); j < nnnn; ++j) cnt[j].second = j;
      ll score = 0;
      for (int j = 0, nnnn = (n); j < nnnn; ++j) {
        if (i & (1 << j))
          score -= arr[j].first;
        else
          score += arr[j].first;
        for (int k = 0, nnnn = (m); k < nnnn; ++k) {
          if (arr[j].second[k] == '1') {
            if (i & (1 << j))
              cnt[k].first++;
            else
              cnt[k].first--;
          }
        }
      }
      sort(cnt, cnt + m);
      for (int j = 0, nnnn = (m); j < nnnn; ++j) {
        score += (j + 1) * cnt[j].first;
      }
      if (ans < score) {
        ans = score;
        for (int j = 0, nnnn = (m); j < nnnn; ++j) {
          p[cnt[j].second] = j + 1;
        }
      }
    }
    for (int i = 0, nnnn = (m); i < nnnn; ++i) {
      cout << p[i] << ' ';
    }
    cout << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout << setprecision(12) << fixed;
    int t;
    cin >> t;
    while (t--) solve();
  }
  