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;
  const long long LINF = 1e17;
  const long double pi = 3.1415926535897932;
  const string endstr = "\n";
  template <typename T>
  T gcd(T a, T b) {
    return (a == 0) ? b : gcd(b % a, a);
  }
  template <typename T>
  T lcm(T a, T b) {
    return a / gcd(a, b) * b;
  }
  bool p_comp_fs(const pair<long long, long long> p1,
                 const pair<long long, long long> p2) {
    return p1.first < p2.first;
  };
  bool p_comp_fg(const pair<long long, long long> p1,
                 const pair<long long, long long> p2) {
    return p1.first > p2.first;
  };
  bool p_comp_ss(const pair<long long, long long> p1,
                 const pair<long long, long long> p2) {
    return p1.second < p2.second;
  };
  bool p_comp_sg(const pair<long long, long long> p1,
                 const pair<long long, long long> p2) {
    return p1.second > p2.second;
  };
  template <typename T>
  vector<T> uniquen(vector<T> vec) {
    sort((vec).begin(), (vec).end());
    vec.erase(unique(vec.begin(), vec.end()), vec.end());
    return vec;
  }
  inline long long popcnt(long long x) {
    return __builtin_popcountll((unsigned long long)x);
  };
  template <class T>
  bool chmax(T &a, T b) {
    if (a < b) {
      a = b;
      return true;
    }
    return false;
  }
  template <class T>
  bool chmin(T &a, T b) {
    if (a > b) {
      a = b;
      return true;
    }
    return false;
  }
  bool bit(long long st, long long b) { return ((st >> b) & 1) == 1; }
  void solve() {
    long long N, M;
    cin >> N >> M;
    vector<long long> X(N);
    for (long long i = 0; i < N; i++) cin >> X[i];
    vector<vector<long long> > S(N, vector<long long>(M, 0));
    for (long long i = 0; i < N; i++)
      for (long long j = 0; j < M; j++) {
        char c;
        cin >> c;
        if (c == '1') S[i][j] = 1;
      }
    long long ans_v = -LINF;
    vector<long long> ans;
    for (long long st = 0; st < (1ll << N); st++) {
      vector<pair<long long, long long> > cnt;
      for (long long j = 0; j < M; j++) {
        long long c = 0;
        for (long long i = 0; i < N; i++) {
          if (bit(st, i))
            c -= S[i][j];
          else
            c += S[i][j];
        }
        cnt.push_back({c, j});
      }
      sort((cnt).begin(), (cnt).end());
      vector<long long> vals(M);
      for (long long j = 0; j < M; j++) {
        vals[cnt[j].second] = j + 1;
      }
      vector<long long> R(N);
      for (long long i = 0; i < N; i++) {
        for (long long j = 0; j < M; j++) {
          R[i] += S[i][j] * vals[j];
        }
      }
      bool ok = true;
      long long ans_c = 0;
      for (long long i = 0; i < N; i++) {
        if (bit(st, i)) {
          ok = ok & (X[i] >= R[i]);
        } else {
          ok = ok & (X[i] <= R[i]);
        }
        ans_c += abs(X[i] - R[i]);
      }
      if (!ok) continue;
      if (chmax(ans_v, ans_c)) {
        ans = vals;
      }
    }
    for (long long v : ans) cout << v << " ";
    cout << endstr;
  }
  int main(int argc, const char *argv[]) {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    long long T;
    cin >> T;
    for (long long _ = 0; _ < T; _++) {
      solve();
    }
    return 0;
  }
  