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;
  constexpr long long int MOD = 998244353;
  constexpr double EPS = 1e-8;
  long long int N, M, K, T, H, W, L, R;
  void Solve() {
    cin >> N >> M;
    vector<int> v(N);
    for (int i = 0; i < N; i++) {
      cin >> v[i];
    }
    vector<string> s(N);
    for (auto& i : s) cin >> i;
    int score = -1;
    vector<int> ans(M);
    for (int i = 0; i < 1 << N; i++) {
      vector<int> sum(N);
      vector<int> point(M);
      int csum = 0;
      vector<pair<int, int>> vp;
      for (int j = 0; j < M; j++) {
        int n = 0;
        for (int k = 0; k < N; k++) {
          if (s[k][j] == '1') {
            if (i >> k & 1) {
              n++;
            } else {
              n--;
            }
          }
        }
        vp.push_back({n, j});
      }
      sort(vp.begin(), vp.end());
      for (int j = 0; j < vp.size(); j++) {
        point[vp[j].second] = j + 1;
      }
      for (int j = 0; j < N; j++) {
        for (int k = 0; k < M; k++) {
          if (s[j][k] == '1') {
            sum[j] += point[k];
          }
        }
      }
      for (int j = 0; j < N; j++) {
        csum += abs(v[j] - sum[j]);
      }
      if (csum > score) {
        score = csum;
        ans = point;
      }
    }
    for (auto i : ans) cout << i << " ";
    cout << endl;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    T = 1;
    cin >> T;
    while (T--) {
      Solve();
    }
  }
  