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;
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int tt;
    cin >> tt;
    while (tt--) {
      int n, m;
      cin >> n >> m;
      vector<int> x(n);
      for (int i = 0; i < n; i++) {
        cin >> x[i];
      }
      vector<vector<int>> a(n, vector<int>(m));
      for (int i = 0; i < n; i++) {
        string s;
        cin >> s;
        for (int j = 0; j < m; j++) {
          a[i][j] = (s[j] == '1');
        }
      }
      const long long inf = (long long)2e18L;
      long long res = -inf, msk = 0;
      for (int i = 0; i < (1 << n); i++) {
        long long z = 0;
        for (int j = 0; j < n; j++) {
          if ((1 << j) & i) {
            z -= x[j];
          } else {
            z += x[j];
          }
        }
        long long ans = z;
        vector<int> cnt(m, 0);
        for (int j = 0; j < n; j++) {
          if ((1 << j) & i) {
            for (int k = 0; k < m; k++) {
              if (a[j][k]) {
                cnt[k] += 1;
              }
            }
          } else {
            for (int k = 0; k < m; k++) {
              if (a[j][k]) {
                cnt[k] -= 1;
              }
            }
          }
        }
        sort(cnt.begin(), cnt.end());
        for (int j = 0; j < m; j++) {
          ans += cnt[j] * 1LL * (j + 1);
        }
        if (ans > res) {
          res = ans;
          msk = i;
        }
      }
      {
        vector<pair<int, int>> cnt(m);
        for (int i = 0; i < m; i++) {
          cnt[i].first = 0;
          cnt[i].second = i;
        }
        for (int j = 0; j < n; j++) {
          if ((1 << j) & msk) {
            for (int k = 0; k < m; k++) {
              if (a[j][k]) {
                cnt[k].first += 1;
              }
            }
          } else {
            for (int k = 0; k < m; k++) {
              if (a[j][k]) {
                cnt[k].first -= 1;
              }
            }
          }
        }
        sort(cnt.begin(), cnt.end());
        vector<int> p(m);
        for (int j = 0; j < m; j++) {
          p[cnt[j].second] = j + 1;
        }
        for (const auto& xx : p) {
          cout << xx << " ";
        }
        cout << '\n';
      }
    }
    return 0;
  }
  