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 = 12;
  const int M = 10010;
  int T, n, m, x[N], w[M], ans[M];
  char s[N][M];
  pair<int, int> a[M];
  int main() {
    scanf("%d", &T);
    for (int t = 1; t <= T; t++) {
      scanf("%d %d", &n, &m);
      for (int i = 0; i < n; i++) {
        scanf("%d", &x[i]);
      }
      for (int i = 0; i < n; i++) {
        scanf("%s", s[i]);
      }
      for (int i = 0; i < m; i++) {
        w[i] = 0;
        for (int j = 0; j < n; j++) {
          w[i] |= ((s[j][i] == '1') ? (1 << j) : 0);
        }
      }
      long long res = -1;
      for (int i = 0; i < (1 << n); i++) {
        int neg = 0;
        int pos = 0;
        long long sum = 0;
        for (int j = 0; j < n; j++) {
          if ((i & (1 << j)) == 0) {
            neg |= (1 << j);
            sum += x[j];
          } else {
            pos |= (1 << j);
            sum -= x[j];
          }
        }
        for (int j = 0; j < m; j++) {
          int mneg = (neg & w[j]);
          int mpos = (pos & w[j]);
          int cnt = __builtin_popcount(mpos) - __builtin_popcount(mneg);
          a[j] = {cnt, j + 1};
        }
        sort(a, a + m);
        for (int j = 0; j < m; j++) {
          sum += a[j].first * (j + 1);
        }
        if (sum > res) {
          res = sum;
          for (int j = 0; j < m; j++) {
            ans[a[j].second - 1] = j + 1;
          }
        }
      }
      for (int i = 0; i < m; i++) {
        printf("%d ", ans[i]);
      }
      puts("");
    }
    return 0;
  }
  