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 target("avx,avx2,fma")
  using namespace std;
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin >> t;
    while (t--) {
      int n, m;
      cin >> n >> m;
      long long exp[n];
      long long score[n];
      int tempscore[m];
      for (int i = 0; i < n; i++) {
        cin >> exp[i];
        score[i] = 0;
      }
      string answers[n];
      for (int i = 0; i < n; i++) {
        cin >> answers[i];
      }
      long long currVal = -1;
      long long bestVal = -1;
      int currPerm[m];
      int bestPerm[m];
      int maxmask = 1 << n;
      for (int mask = 0; mask < maxmask; mask++) {
        memset(tempscore, 0, sizeof(tempscore));
        for (int i = 0; i < n; i++) score[i] = 0;
        currVal = 0;
        for (int i = 0; i < n; i++) {
          if ((mask >> i) & 1) {
            for (int j = 0; j < m; j++) {
              tempscore[j] += answers[i][j] == '1';
            }
          } else {
            for (int j = 0; j < m; j++) {
              tempscore[j] -= answers[i][j] == '1';
            }
          }
        }
        vector<pair<int, int> > myVec;
        for (int j = 0; j < m; j++) {
          myVec.push_back({tempscore[j], j});
        }
        sort(myVec.begin(), myVec.end());
        for (int j = 0; j < m; j++) {
          currPerm[myVec[j].second] = j + 1;
          for (int i = 0; i < n; i++) {
            score[i] += (answers[i][myVec[j].second] == '1') * (j + 1);
          }
        }
        for (int i = 0; i < n; i++) {
          currVal += abs(score[i] - exp[i]);
        }
        if (currVal > bestVal) {
          bestVal = currVal;
          for (int j = 0; j < m; j++) {
            bestPerm[j] = currPerm[j];
          }
        }
      }
      for (int j = 0; j < m; j++) {
        cout << bestPerm[j] << " ";
      }
      cout << "\n";
    }
  }
  