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() {
    int t;
    scanf("%d", &t);
    while (t--) {
      int n, m;
      scanf("%d%d", &n, &m);
      vector<int> x(n);
      for (int i = 0; i < int(n); ++i) scanf("%d", &x[i]);
      vector<vector<int>> a(n, vector<int>(m));
      for (int i = 0; i < int(n); ++i)
        for (int j = 0; j < int(m); ++j) scanf("%1d", &a[i][j]);
      int ans = -1;
      vector<int> best;
      for (int mask = 0; mask < int(1 << n); ++mask) {
        vector<int> val(m);
        for (int i = 0; i < int(n); ++i)
          for (int j = 0; j < int(m); ++j)
            if (a[i][j]) val[j] += ((mask >> i) & 1) ? +1 : -1;
        int res = 0;
        for (int i = 0; i < int(n); ++i) res += ((mask >> i) & 1) ? -x[i] : x[i];
        vector<int> p(m);
        iota(p.begin(), p.end(), 0);
        sort(p.begin(), p.end(), [&](int x, int y) { return val[x] < val[y]; });
        for (int i = 0; i < int(m); ++i) res += val[p[i]] * (i + 1);
        if (res > ans) ans = res, best = p;
      }
      vector<int> ansPerm(m);
      for (int i = 0; i < int(m); ++i) ansPerm[best[i]] = i;
      for (int i = 0; i < int(m); ++i) printf("%d ", ansPerm[i] + 1);
      puts("");
    }
  }
  