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;
  char s[10][10001];
  int X[10];
  pair<int, vector<int>> Calculate(int n, int m, int mask) {
    vector<pair<int, int>> v(m);
    for (int k = 0; k < m; ++k) {
      int c = 0;
      for (int i = 0; i < n; ++i) {
        if (s[i][k] == '0') {
          continue;
        }
        c += ((mask & (1 << i)) != 0) ? 1 : -1;
      }
      v[k] = {c, k};
    }
    int ans = 0;
    for (int i = 0; i < n; ++i) {
      int c = ((mask & (1 << i)) != 0) ? -1 : 1;
      ans += c * X[i];
    }
    vector<int> p(m);
    sort(v.begin(), v.end());
    for (int k = 0; k < m; ++k) {
      ans += v[k].first * (k + 1);
      p[v[k].second] = k + 1;
    }
    return {ans, p};
  }
  void Solve() {
    int n, m;
    scanf("%d%d", &n, &m);
    for (int i = 0; i < n; ++i) {
      scanf("%d", &X[i]);
    }
    for (int i = 0; i < n; ++i) {
      scanf("%s", s[i]);
    }
    pair<int, vector<int>> ans = {INT_MIN, vector<int>()};
    for (int mask = 0; mask < (1 << n); ++mask) {
      ans = max(ans, Calculate(n, m, mask));
    }
    auto [_, p] = ans;
    for (int pi : p) {
      printf("%d ", pi);
    }
    printf("\n");
  }
  int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
      Solve();
    }
    return 0;
  }
  