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 optimize("unroll-loops")
  using namespace std;
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  int n, m;
  long long res = -1;
  string t[20];
  long long w[20];
  int ans[200000];
  void get(long long x) {
    int s = 0;
    int k[n], f[n], p = 0;
    for (int i = 0; i < n; i++) k[i] = 1;
    while (x > 0) {
      if (x % 2 == 1) {
        k[p] = -1;
        s += w[p];
      }
      p++;
      x /= 2;
    }
    for (int i = 0; i < n; i++)
      if (k[i] == 1) s -= w[i];
    vector<pair<int, int> > a(m);
    for (int i = 0; i < m; i++) {
      a[i].first = 0;
      a[i].second = i;
    }
    for (int i = 0; i < n; i++)
      for (int j = 0; j < m; j++)
        if (t[i][j] == '1') a[j].first += k[i];
    sort(a.begin(), a.end());
    for (int i = 0; i < m; i++) s += a[i].first * (i + 1);
    if (s > res) {
      res = s;
      for (int i = 0; i < m; i++) ans[a[i].second] = i + 1;
      return;
    }
  }
  void solve() {
    cin >> n >> m;
    for (int i = 0; i < n; i++) cin >> w[i];
    for (int i = 0; i < n; i++) cin >> t[i];
    res = -1;
    for (int x = 0; x < (1 << n); x++) get(x);
    for (int i = 0; i < m; i++) cout << ans[i] << ' ';
    cout << "\n";
  }
  int main() {
    ios::sync_with_stdio(0);
    int q;
    cin >> q;
    while (q--) solve();
  }
  