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;
  using ll = long long;
  using pii = pair<int, int>;
  const int maxn = 1e4 + 10;
  int n, m;
  int a[maxn];
  char s[12][maxn];
  void solve() {
    scanf("%d%d", &n, &m);
    for (int i = 0; i < n; i++) scanf("%d", &a[i]);
    for (int i = 0; i < n; i++) {
      scanf("%s", s[i]);
    }
    ll ans = -1e18;
    vector<int> p(m);
    for (int b = 0; b < 1 << n; b++) {
      vector<int> f(n);
      ll now = 0;
      for (int i = 0; i < n; i++) {
        if (b >> i & 1)
          f[i] = 1;
        else
          f[i] = -1;
        if (f[i] == -1)
          now += a[i];
        else
          now -= a[i];
      }
      vector<pii> cost(m);
      for (int i = 0; i < m; i++) cost[i].second = i;
      for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
          if (s[i][j] != '0') cost[j].first += f[i];
        }
      }
      sort(cost.begin(), cost.end());
      for (int i = 0; i < m; i++) {
        now += 1ll * cost[i].first * (i + 1);
      }
      if (now >= ans) {
        ans = now;
        for (int i = 0; i < m; i++) {
          p[cost[i].second] = i + 1;
        }
      }
    }
    for (int i = 0; i < m; i++) printf("%d%c", p[i], " \n"[i == m - 1]);
  }
  int main() {
    int t;
    cin >> t;
    while (t--) solve();
  }
  