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("O3")
  #pragma GCC target("popcnt")
  using namespace std;
  mt19937 bruh(chrono::steady_clock::now().time_since_epoch().count());
  const long long N = 1e6 + 100;
  const long long mod = 998244353;
  const long long big = 1e17;
  long long t, n, m;
  string a[N];
  pair<long long, long long> c[N];
  int x[N], d[N], e[N];
  int main() {
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cin >> t;
    while (t--) {
      cin >> n >> m;
      for (int i = 0; i < n; i++) cin >> x[i];
      for (int i = 0; i < n; i++) cin >> a[i];
      long long ans = -1e10;
      for (int j = 0; j < (1 << n); j++) {
        for (int i = 0; i < m; i++) c[i] = make_pair(0, i), d[i] = 0;
        long long sum = 0;
        for (int i = 0; i < n; i++) {
          if ((1 << i) & j) {
            sum += x[i];
            for (int k = 0; k < m; k++) {
              if (a[i][k] == '1') c[k].first--;
            }
          } else {
            sum -= x[i];
            for (int k = 0; k < m; k++)
              if (a[i][k] == '1') c[k].first++;
          }
        }
        sort(c, c + m);
        for (int i = 0; i < m; i++)
          d[c[i].second] = i + 1, sum = sum + c[i].first * (i + 1);
        bool t = 1;
        for (int i = 0; i < n; i++) {
          long long sum2 = 0;
          for (int k = 0; k < m; k++)
            if (a[i][k] == '1') sum2 += d[k];
          if (((1 << i) & j) && sum2 > x[i]) t = 0;
          if ((!((1 << i) & j)) && sum2 < x[i]) t = 0;
        }
        if (t && (sum > ans)) {
          ans = sum;
          for (int j = 0; j < m; j++) e[j] = d[j];
        }
      }
      for (int i = 0; i < m; i++) cout << e[i] << " ";
      cout << '\n';
    }
  }
  