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>
  std::mt19937 rng(
      (unsigned int)std::chrono::steady_clock::now().time_since_epoch().count());
  using namespace std;
  void solve(int ttt) {
    long long int n, m;
    cin >> n >> m;
    vector<long long int> x(n);
    for (long long int i = 0; i < n; i++) cin >> x[i];
    vector<vector<int> > a(n, vector<int>(m));
    for (long long int i = 0; i < n; i++) {
      string s;
      cin >> s;
      for (long long int j = 0; j < m; j++) {
        if (s[j] == '1')
          a[i][j] = 1;
        else
          a[i][j] = 0;
      }
    }
    long long int res = LLONG_MIN;
    vector<long long int> perm;
    for (int msk = 0; msk < (1 << n); msk++) {
      vector<int> coeff(m, 0);
      long long int curVal = 0;
      for (long long int i = 0; i < n; i++) {
        if (msk & (1 << i)) {
          curVal += x[i];
          for (long long int j = 0; j < m; j++) {
            if (a[i][j]) coeff[j]--;
          }
        } else {
          curVal -= x[i];
          for (long long int j = 0; j < m; j++) {
            if (a[i][j]) coeff[j]++;
          }
        }
      }
      vector<int> ind(m);
      iota(ind.begin(), ind.end(), 0);
      sort(ind.begin(), ind.end(),
           [&](int i, int j) { return coeff[i] < coeff[j]; });
      long long int c = 1;
      for (auto i : ind) {
        curVal += c * coeff[i];
        c++;
      }
      if (res < curVal) {
        res = curVal;
        vector<long long int> tmp(m);
        for (long long int i = 0; i < m; i++) {
          tmp[ind[i]] = i + 1;
        }
        swap(perm, tmp);
      }
    }
    for (long long int i = 0; i < m; i++) {
      cout << perm[i] << " ";
    }
    cout << "\n";
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int ttt;
    ttt = 1;
    cin >> ttt;
    for (long long int i = 0; i < ttt; i++) solve(i);
  }
  