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;
  const int N = 1e5 + 5;
  const int M = 35;
  int x[M];
  int a[M][N];
  struct ans {
    int A, B;
  } cnt[N];
  bool cmp(ans C, ans D) {
    if (C.A == D.A)
      return C.B < D.B;
    else
      return C.A < D.A;
  }
  int res[N];
  void run() {
    int n, m;
    scanf("%d%d", &n, &m);
    for (int i = 0; i < n; i++) {
      scanf("%d", &x[i]);
    }
    for (int i = 0; i < m; i++) cnt[i].A = 0, cnt[i].B = i + 1;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        char c;
        cin >> c;
        if (c == '1')
          a[i][j] = 1;
        else
          a[i][j] = 0;
      }
    }
    int nm = (1 << n);
    long long ans = -1e18;
    for (int i = 0; i < nm; i++) {
      long long tot = 0;
      for (int j = 0; j < m; j++) cnt[j].A = 0, cnt[j].B = j + 1;
      for (int j = 0; j < n; j++) {
        if (i >> j & 1)
          tot += x[j];
        else
          tot -= x[j];
      }
      for (int j = 0; j < n; j++) {
        for (int k = 0; k < m; k++) {
          if ((i >> j & 1) && a[j][k])
            cnt[k].A--;
          else if (!(i >> j & 1) && a[j][k])
            cnt[k].A++;
        }
      }
      sort(cnt, cnt + m, cmp);
      for (int j = 0; j < m; j++) tot += cnt[j].A * (j + 1);
      if (tot > ans) {
        ans = tot;
        for (int j = 0; j < m; j++) {
          res[cnt[j].B] = j + 1;
        }
      }
    }
    for (int i = 1; i <= m; i++) {
      printf("%d ", res[i]);
    }
    puts("");
  }
  int main() {
    int T;
    scanf("%d", &T);
    while (T--) {
      run();
    }
    return 0;
  }
  