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 main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while (t--) {
      int n, m;
      cin >> n >> m;
      vector<int> a(n);
      for (int i = 0; i < n; ++i) cin >> a[i];
      vector<string> b(n);
      for (int i = 0; i < n; ++i) cin >> b[i];
      vector<int> ans;
      long long mx = -1;
      for (int i = 0; i < 1 << n; ++i) {
        long long surprise = 0;
        vector<int> weight(m);
        for (int j = 0; j < n; ++j) {
          if (i >> j & 1) {
            for (int k = 0; k < m; ++k) {
              if (b[j][k] == '1') weight[k] += 1;
            }
            surprise -= a[j];
          } else {
            for (int k = 0; k < m; ++k) {
              if (b[j][k] == '1') weight[k] -= 1;
            }
            surprise += a[j];
          }
        }
        vector<int> ord(m), r(m);
        iota(ord.begin(), ord.end(), 0);
        sort(ord.begin(), ord.end(),
             [&](int i, int j) { return weight[i] < weight[j]; });
        for (int j = 0; j < m; ++j) {
          r[ord[j]] = j;
          surprise += (j + 1) * weight[ord[j]];
        }
        if (surprise > mx) {
          mx = surprise;
          ans = r;
        }
      }
      for (int i : ans) {
        cout << i + 1 << " ";
      }
      cout << '\n';
    }
  }
  