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 long long N = 15, M = 1e4 + 1e2 + 7;
  long long T;
  long long n, m, x[N];
  char s[N][M];
  long long val[M], r[M], p[M], ap[M];
  bool cmp(long long a, long long b) { return val[a] < val[b]; }
  signed main() {
    scanf("%lld", &T);
    while (T--) {
      long long ans = -1;
      scanf("%lld%lld", &n, &m);
      for (long long i = 1; i <= n; i++) scanf("%lld", &x[i]);
      for (long long i = 1; i <= n; i++) scanf("%s", s[i] + 1);
      for (long long S = 0; S < (1 << n); S++) {
        for (long long i = 1; i <= m; i++) val[i] = 0, r[i] = i;
        for (long long i = 1; i <= n; i++) {
          long long v = (S & (1 << i - 1)) ? 1 : -1;
          for (long long j = 1; j <= m; j++)
            if (s[i][j] == '1') val[j] += v;
        }
        sort(r + 1, r + m + 1, cmp);
        for (long long i = 1; i <= m; i++) p[r[i]] = i;
        long long tot = 0;
        for (long long i = 1; i <= n; i++) {
          long long sv = 0;
          for (long long j = 1; j <= m; j++)
            if (s[i][j] == '1') sv += p[j];
          tot += abs(sv - x[i]);
        }
        if (tot > ans) {
          ans = tot;
          for (long long j = 1; j <= m; j++) ap[j] = p[j];
        }
      }
      for (long long i = 1; i <= m; i++) printf("%lld%c", ap[i], " \n"[i == m]);
    }
  }
  