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;
  int x[10], re[10001], anss[10001];
  string s[10];
  pair<int, int> h[10001];
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    int T;
    cin >> T;
    while (T--) {
      int n, m;
      cin >> n >> m;
      for (int i = 0; i < n; i++) cin >> x[i];
      for (int i = 0; i < n; i++) cin >> s[i];
      vector<vector<int>> b(m + 1);
      for (int i = 1; i <= m; i++)
        for (int j = 0; j < n; j++)
          if (s[j][i - 1] == '1') b[i].push_back(j);
      int ans = INT_MIN;
      for (int bm = 0; bm < (1 << n); bm++) {
        for (int i = 1; i <= m; i++) {
          h[i].first = 0;
          for (int &j : b[i])
            h[i].first +=
                ((bm >> j) & 1) ? (s[j][i - 1] - '0') : ('0' - s[j][i - 1]);
          h[i].second = i;
        }
        int t = 0;
        for (int i = 0; i < n; i++) t += ((bm >> i) & 1 ? -x[i] : x[i]);
        sort(h + 1, h + m + 1);
        for (int i = 1; i <= m; i++) t += h[i].first * i;
        if (t > ans) {
          ans = t;
          for (int i = 1; i <= m; i++) anss[h[i].second] = i;
        }
      }
      for (int i = 1; i <= m; i++) cout << anss[i] << ' ';
      cout << '\n';
    }
    return 0;
  }
  