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;
  const int MX = 10005;
  int n, m, x[11], acu[MX], cn[22];
  long long sum[MX];
  string s[11];
  vector<int> res;
  void main_() {
    cin >> n >> m;
    for (int i = 0; i < int(n); i++) cin >> x[i];
    for (int i = 0; i < int(n); i++) cin >> s[i];
    long long mx = -1e18;
    int rm;
    for (int mk = 0; mk < int(1 << n); mk++) {
      memset(acu, 0, sizeof(acu));
      long long p = 0;
      for (int i = 0; i < int(n); i++) {
        int d = 0;
        if (mk & (1 << i)) {
          p -= x[i];
          d = 1;
        } else {
          p += x[i];
          d = -1;
        }
        for (int j = 0; j < int(m); j++)
          if (s[i][j] == '1') acu[j] += d;
      }
      memset(cn, 0, sizeof(cn));
      for (int j = 0; j < int(m); j++) cn[acu[j] + 10]++;
      for (int i = 20, x = 10, t = m; i >= 0; i--, x--) {
        p += x * sum[t];
        t -= cn[i];
        p -= x * sum[t];
      }
      if (mx < p) {
        mx = p;
        rm = mk;
      }
    }
    vector<pair<int, int> > v(m);
    for (int i = 0; i < int(m); i++) v[i].second = i;
    for (int i = 0; i < int(n); i++) {
      int d = (rm & (1 << i)) ? 1 : -1;
      for (int j = 0; j < int(m); j++)
        if (s[i][j] == '1') v[j].first += d;
    }
    sort(v.begin(), v.end());
    vector<int> res(m);
    for (int i = 0; i < int(m); i++) res[v[i].second] = i + 1;
    for (int i = 0; i < int(m); i++) cout << res[i] << " ";
    cout << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    for (int i = 0; i < int(MX); i++) {
      sum[i] = i;
      if (i) sum[i] += sum[i - 1];
    }
    int t = 1;
    cin >> t;
    while (t--) main_();
    return 0;
  }
  