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;
  using ll = long long;
  constexpr int INF = 0x3f3f3f3f;
  constexpr long long LINF = 0x3f3f3f3f3f3f3f3fLL;
  constexpr double EPS = 1e-8;
  constexpr int MOD = 1000000007;
  constexpr int DY[]{1, 0, -1, 0}, DX[]{0, -1, 0, 1};
  constexpr int DY8[]{1, 1, 0, -1, -1, -1, 0, 1},
      DX8[]{0, -1, -1, -1, 0, 1, 1, 1};
  template <typename T, typename U>
  inline bool chmax(T& a, U b) {
    return a < b ? (a = b, true) : false;
  }
  template <typename T, typename U>
  inline bool chmin(T& a, U b) {
    return a > b ? (a = b, true) : false;
  }
  struct IOSetup {
    IOSetup() {
      std::cin.tie(nullptr);
      std::ios_base::sync_with_stdio(false);
      std::cout << fixed << setprecision(20);
    }
  } iosetup;
  void solve() {
    int n, m;
    cin >> n >> m;
    vector<int> x(n);
    for (int i = (0); i < (n); ++i) cin >> x[i];
    vector<string> s(n);
    for (int i = (0); i < (n); ++i) cin >> s[i];
    ll surprise = -LINF;
    vector<int> ans;
    for (int bit = (0); bit < (1 << n); ++bit) {
      ll value = 0;
      for (int i = (0); i < (n); ++i) {
        if (bit >> i & 1) {
          value += x[i];
        } else {
          value -= x[i];
        }
      }
      vector<int> coef(m, 0);
      for (int j = (0); j < (m); ++j) {
        for (int i = (0); i < (n); ++i) {
          if (s[i][j] == '1') {
            if (bit >> i & 1) {
              --coef[j];
            } else {
              ++coef[j];
            }
          }
        }
      }
      vector<int> ord(m);
      iota((ord).begin(), (ord).end(), 0);
      sort((ord).begin(), (ord).end(),
           [&](int l, int r) -> bool { return coef[l] < coef[r]; });
      vector<int> p(m);
      for (int i = (0); i < (m); ++i) {
        p[ord[i]] = i + 1;
        value += coef[ord[i]] * (i + 1);
      }
      if (chmax(surprise, value)) ans = p;
    }
    for (int i = (0); i < (m); ++i) cout << ans[i] << " \n"[i + 1 == m];
  }
  int main() {
    int t;
    cin >> t;
    while (t--) solve();
    return 0;
  }
  