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);
    long long t;
    cin >> t;
    while (t--) {
      long long n, m;
      cin >> n >> m;
      vector<long long> a(n);
      for (int i = 0, ggdem = n; i < ggdem; ++i) cin >> a[i];
      vector<string> s(n);
      for (int i = 0, ggdem = n; i < ggdem; ++i) cin >> s[i];
      long long maxi = -1;
      vector<long long> res(m);
      for (int i = 0, ggdem = (1ll << n); i < ggdem; ++i) {
        vector<pair<long long, long long>> c(m);
        for (int j = 0, ggdem = n; j < ggdem; ++j) {
          long long p = 1;
          if (i & (1ll << j)) p = -1;
          for (int ii = 0, ggdem = m; ii < ggdem; ++ii) {
            if (s[j][ii] == '1') c[ii].first += p;
          }
        }
        for (int ii = 0, ggdem = m; ii < ggdem; ++ii) c[ii].second = ii;
        sort(c.begin(), c.end());
        long long val = 0;
        for (int ii = 0, ggdem = m; ii < ggdem; ++ii)
          val += (ii + 1) * c[ii].first;
        for (int j = 0, ggdem = n; j < ggdem; ++j) {
          if (i & (1ll << j))
            val += a[j];
          else
            val -= a[j];
        }
        if (val > maxi) {
          maxi = val;
          for (int ii = 0, ggdem = m; ii < ggdem; ++ii) res[c[ii].second] = ii;
        }
      }
      for (auto i : res) cout << i + 1 << " ";
      cout << "\n";
    }
    return 0;
  }
  