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>
  #pragma GCC optimize("Ofast")
  #pragma GCC optimization("unroll-loops, no-stack-protector")
  #pragma GCC target("avx,avx2,fma")
  using namespace std;
  using pii = pair<int, int>;
  using pll = pair<long long, long long>;
  using vb = vector<bool>;
  using vi = vector<int>;
  using vl = vector<long long>;
  using vpii = vector<pair<int, int>>;
  using vvb = vector<vector<bool>>;
  using vvi = vector<vector<int>>;
  using vvl = vector<vector<long long>>;
  using vvpii = vector<vector<pair<int, int>>>;
  using mii = map<int, int>;
  using umii = unordered_map<int, int>;
  using seti = set<int>;
  using useti = unordered_set<int>;
  vpii cost(10000);
  vi inv(10000);
  vi perm(10000);
  inline void __run_test() {
    int n, m;
    cin >> n >> m;
    vi x(n);
    vector<string> d(n);
    for (int i = 0; i < (n); i++) cin >> x[i];
    for (int i = 0; i < (n); i++) cin >> d[i];
    long long val = -1;
    for (int i = 0; i < (m); i++) cost[i].second = i;
    for (int mask = 0; mask < (1 << n); mask++) {
      for (int i = 0; i < (m); i++) cost[i].first = 0;
      sort(cost.begin(), cost.begin() + m,
           [&](pii a, pii b) -> bool { return a.second < b.second; });
      for (int i = 0; i < (n); i++)
        for (int j = 0; j < (m); j++)
          if (d[i][j] == '1') cost[j].first += (mask & (1 << i)) ? 1 : -1;
      sort(cost.begin(), cost.begin() + m);
      for (int i = 0; i < (m); i++) inv[cost[i].second] = i + 1;
      long long now = 0;
      for (int i = 0; i < (n); i++) {
        long long r = 0;
        for (int j = 0; j < (m); j++)
          if (d[i][j] == '1') r += inv[j];
        now += abs(x[i] - r);
      }
      if (now > val) {
        val = now;
        for (int i = 0; i < (m); i++) perm[i] = inv[i];
      }
    }
    for (int i = 0; i < (m); i++) cout << perm[i] << " ";
    cout << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int __tests;
    cin >> __tests;
    while (__tests--) __run_test();
  }
  