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;
  long long a[15];
  int k[10005];
  char s[11][10005];
  vector<long long> v[200005], vv, save;
  int outs[20005];
  int main() {
    long long ans;
    int t;
    cin >> t;
    while (t--) {
      ans = -1e18;
      int n, m;
      cin >> n >> m;
      for (int i = 0; i < n; i++) scanf("%lld", a + i);
      for (int i = 0; i < n; i++) scanf("%s", s[i]);
      for (int i = 0; i < (1 << n); i++) {
        for (int j = 0; j <= 20; j++) v[j].clear();
        vv.clear();
        long long c = 0;
        for (int j = 0; j < n; j++) {
          if (i & (1 << j))
            c += a[j];
          else
            c -= a[j];
        }
        for (int j = 0; j < m; j++) k[j] = 0;
        for (int j = 0; j < n; j++) {
          if (i & (1 << j)) {
            for (int q = 0; q < m; q++)
              if (s[j][q] == '1') k[q]--;
          } else {
            for (int q = 0; q < m; q++)
              if (s[j][q] == '1') k[q]++;
          }
        }
        for (int j = 0; j < m; j++) {
          v[k[j] + 10].push_back(j);
        }
        for (int j = 0; j <= 20; j++)
          for (auto &k : v[j]) vv.push_back(k);
        for (int j = 0; j < m; j++) {
          c += k[vv[j]] * (j + 1);
        }
        if (c > ans) {
          ans = c;
          save = vv;
        }
      }
      int cnt = 0;
      for (auto k : save) {
        outs[k] = ++cnt;
      }
      for (int i = 0; i < m; i++) printf("%d%c", outs[i], " \n"[i == m - 1]);
    }
    return 0;
  }
  