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>
  namespace {
  using namespace std;
  template <class T>
  void chmax(T& a, const T& b) {
    a = max(a, b);
  }
  template <class T>
  void chmin(T& a, const T& b) {
    a = min(a, b);
  }
  using ll = long long;
  using P = pair<int, int>;
  using VI = vector<int>;
  using VVI = vector<VI>;
  using VL = vector<ll>;
  using VVL = vector<VL>;
  bool s[10][10000];
  int x[10];
  int tmp[10000];
  int hist[21];
  int ans[10000];
  int ptr[21];
  }  // namespace
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int tt;
    cin >> tt;
    while (tt--) {
      int n, m;
      cin >> n >> m;
      for (int i = 0; i < int(n); ++i) cin >> x[i];
      for (int i = 0; i < int(n); ++i)
        for (int j = 0; j < int(m); ++j) {
          char c;
          cin >> c;
          s[i][j] = c == '1';
        }
      ll mx = -1;
      for (int b = 0; b < int(1 << n); ++b) {
        memset(tmp, 0, sizeof(int) * m);
        ll sm = 0;
        for (int i = 0; i < int(n); ++i) {
          if (b >> i & 1) {
            sm += x[i];
            for (int j = 0; j < int(m); ++j) tmp[j] -= s[i][j];
          } else {
            sm -= x[i];
            for (int j = 0; j < int(m); ++j) tmp[j] += s[i][j];
          }
        }
        memset(hist, 0, sizeof(int) * 21);
        for (int j = 0; j < int(m); ++j) hist[tmp[j] + 10]++;
        ptr[0] = 0;
        for (int i = 0; i < int(20); ++i) ptr[i + 1] = ptr[i] + hist[i];
        for (int j = 0; j < int(m); ++j) {
          sm += tmp[j] * (ptr[tmp[j] + 10]++ + 1);
        }
        if (sm > mx) {
          mx = sm;
          ptr[0] = 0;
          for (int i = 0; i < int(20); ++i) ptr[i + 1] = ptr[i] + hist[i];
          for (int j = 0; j < int(m); ++j) ans[j] = ptr[tmp[j] + 10]++ + 1;
        }
      }
      for (int j = 0; j < int(m); ++j) cout << ans[j] << " \n"[j + 1 == m];
    }
  }
  