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,no-stack-protector,unroll-loops")
  #pragma GCC target("avx2,fma")
  using namespace std;
  const long long inf = 1e18, N = 1e4 + 10, maxn = 11, mod = 1e9 + 7,
                  pr = 1000696969;
  int x[maxn], st[N], dp[N];
  vector<pair<long long, int> > vc;
  string s[maxn];
  inline void solve() {
    int n, m;
    cin >> n >> m;
    for (int i = 0; i < n; i++) cin >> x[i];
    for (int i = 0; i < n; i++) cin >> s[i];
    long long ans = 0, ans_t = -inf;
    for (int mask = 0; mask < (1 << n); mask++) {
      vc.clear();
      ans = 0;
      for (int j = 0; j < m; j++) dp[j] = 0;
      for (int j = 0; j < n; j++) {
        if (mask & (1 << j)) {
          ans += x[j];
          for (int k = 0; k < m; k++) {
            if (s[j][k] == '1') {
              dp[k]--;
            }
          }
        } else {
          ans -= x[j];
          for (int k = 0; k < m; k++) {
            if (s[j][k] == '1') {
              dp[k]++;
            }
          }
        }
      }
      for (int i = 0; i < m; i++) {
        vc.push_back({dp[i], i});
      }
      sort((vc).begin(), (vc).end());
      for (int i = int((vc).size()) - 1; i >= 0; i--) {
        ans += 1ll * vc[i].first * (i + 1);
      }
      if (ans > ans_t) {
        for (int i = int((vc).size()) - 1; i >= 0; i--) {
          st[vc[i].second] = i + 1;
        }
        ans_t = ans;
      }
    }
    for (int i = 0; i < m; i++) cout << st[i] << " ";
    cout << "\n";
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    ;
    int t;
    cin >> t;
    while (t--) solve();
    return 0;
  }
  