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 n, m;
  int a[10], b[10][10005];
  void Solve();
  int main() {
    int tests = 1;
    scanf("%d", &tests);
    while (tests--) {
      Solve();
    }
    return 0;
  }
  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)
      for (int j = 1; j <= (m); ++j) scanf("%1d", &b[i][j]);
    static int ans[10005];
    int maxdif = -1;
    for (int s = 0; s < (1 << n); ++s) {
      static int c[10005];
      for (int j = 1; j <= (m); ++j) c[j] = 0;
      for (int i = 0; i < (n); ++i)
        for (int j = 1; j <= (m); ++j) c[j] += b[i][j] ? s >> i & 1 ? -1 : 1 : 0;
      static int per[10005], val[10005];
      for (int j = 1; j <= (m); ++j) per[j] = j;
      sort(per + 1, per + m + 1, [&](int i, int j) { return c[i] < c[j]; });
      for (int j = 1; j <= (m); ++j) val[per[j]] = j;
      int dif = 0;
      for (int i = 0; i < (n); ++i) {
        int sc = 0;
        for (int j = 1; j <= (m); ++j)
          if (b[i][j]) sc += val[j];
        dif += abs(a[i] - sc);
      }
      if (dif > maxdif) {
        maxdif = dif;
        for (int j = 1; j <= (m); ++j) ans[j] = val[j];
      }
    }
    for (int j = 1; j <= (m); ++j) printf("%d%c", ans[j], " \n"[j == m]);
  }
  