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, a[10], c[10005];
  char b[10][10005];
  vector<int> g[10];
  int id[10005];
  bool cmp(int i1, int i2) { return c[i1] < c[i2]; }
  int p[10005], p2[10005];
  int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
      scanf("%d%d", &n, &m);
      for (int i = 0; i < n; i++) {
        scanf("%d", &a[i]);
      }
      for (int i = 0; i < n; i++) {
        scanf("\n%s", b[i] + 1);
        g[i].clear();
        for (int j = 1; j <= m; j++) {
          if (b[i][j] == '1') {
            g[i].push_back(j);
          }
        }
      }
      long long ans = 0;
      for (int i = 1; i <= m; i++) id[i] = i;
      for (int i = 0; i < (1 << n); i++) {
        memset(c, 0, sizeof(c));
        for (int j = 0; j < n; j++) {
          for (int k : g[j]) {
            if (i & (1 << j))
              c[k]++;
            else
              c[k]--;
          }
        }
        sort(id + 1, id + 1 + m, cmp);
        for (int j = 1; j <= m; j++) {
          p[id[j]] = j;
        }
        long long sum = 0;
        for (int j = 0; j < n; j++) {
          int h = 0;
          for (int k : g[j]) {
            h += p[k];
          }
          sum += abs(a[j] - h);
        }
        if (sum >= ans) {
          ans = sum;
          for (int j = 1; j <= m; j++) {
            p2[j] = p[j];
          }
        }
      }
      for (int i = 1; i <= m; i++) {
        printf("%d ", p2[i]);
      }
      printf("\n");
    }
    return 0;
  }
  