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;
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    long long t;
    cin >> t;
    while (t--) {
      long long n, m;
      cin >> n >> m;
      vector<long long> x(n);
      for (long long i = 0; i < n; ++i) {
        cin >> x[i];
      }
      vector<string> s(n);
      for (long long i = 0; i < n; ++i) {
        cin >> s[i];
      }
      vector<vector<long long>> gr(1 << n);
      for (long long j = 0; j < m; ++j) {
        long long sum = 0;
        for (long long i = 0; i < n; ++i) {
          sum = (sum << 1) + s[i][j] - '0';
        }
        gr[sum].push_back(j);
      }
      vector<vector<long long>> srt;
      for (long long i = 0; i < (1 << n); ++i) {
        if ((long long)gr[i].size()) {
          srt.push_back(gr[i]);
        }
      }
      vector<long long> col(m);
      for (long long i = 0; i < (long long)srt.size(); ++i) {
        for (auto& j : srt[i]) {
          col[j] = i;
        }
      }
      vector<vector<long long>> uses(n);
      for (long long i = 0; i < n; ++i) {
        for (long long j = 0; j < m; ++j) {
          if (s[i][j] == '1') {
            uses[i].push_back(col[j]);
          }
        }
        sort(uses[i].begin(), uses[i].end());
        uses[i].erase(unique(uses[i].begin(), uses[i].end()), uses[i].end());
      }
      long long ans = 0;
      vector<long long> vans;
      for (long long i = 0; i < (1 << n); ++i) {
        long long sum = 0;
        vector<long long> gcnt((long long)srt.size());
        for (long long j = 0; j < n; ++j) {
          if (i & (1 << j)) {
            sum += x[j];
            for (auto& x : uses[j]) {
              --gcnt[x];
            }
          } else {
            sum -= x[j];
            for (auto& x : uses[j]) {
              ++gcnt[x];
            }
          }
        }
        vector<vector<long long>> gsrt;
        for (long long h = 0; h < (long long)srt.size(); ++h) {
          gsrt.push_back({gcnt[h], (long long)srt[h].size(), h});
        }
        sort(gsrt.begin(), gsrt.end());
        long long l = 1;
        for (long long h = 0; h < (long long)srt.size(); ++h) {
          long long r = l + (long long)gsrt[h][1] - 1;
          sum += (r + l) * (r - l + 1) / 2 * gsrt[h][0];
          l = r + 1;
        }
        if (sum >= ans) {
          ans = sum;
          vans.clear();
          for (long long h = 0; h < (long long)srt.size(); ++h) {
            vans.push_back(gsrt[h][2]);
          }
        }
      }
      vector<long long> out(m);
      long long l = 1;
      for (auto& i : vans) {
        for (auto& j : srt[i]) {
          out[j] = l++;
        }
      }
      for (long long i = 0; i < m; ++i) {
        cout << out[i] << ' ';
      }
      cout << '\n';
    }
    return 0;
  }
  