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--) {
      static int xx[10], ans[10000];
      static char s[10][10000 + 1];
      int n, m, ansv;
      scanf("%d%d", &n, &m);
      for (int i = 0; i < n; i++) scanf("%d", &xx[i]);
      for (int i = 0; i < n; i++) scanf("%s", s[i]);
      ansv = -1;
      memset(ans, 0, m * sizeof *ans);
      for (int b = 0; b < 1 << n; b++) {
        static int kk[10000];
        int sum = 0, cnt;
        vector<int> ii(m);
        for (int i = 0; i < n; i++)
          if (b >> i & 1)
            sum += xx[i];
          else
            sum -= xx[i];
        for (int h = 0; h < m; h++) {
          int k = 0;
          for (int i = 0; i < n; i++)
            if (s[i][h] == '1') {
              if (b >> i & 1)
                k--;
              else
                k++;
            }
          kk[h] = k, ii[h] = h;
        }
        sort(ii.begin(), ii.end(), [&](int i, int j) { return kk[i] < kk[j]; });
        cnt = 1;
        for (int h : ii) sum += kk[h] * cnt++;
        if (sum > ansv) {
          ansv = sum;
          cnt = 1;
          for (int h : ii) {
            ans[h] = cnt++;
          }
        }
      }
      for (int h = 0; h < m; h++) printf("%d ", ans[h]);
      printf("\n");
    }
    return 0;
  }
  