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 MAXN = 10 + 5;
  const int MAXM = 1e4 + 5;
  const long long linf = 0x3f3f3f3f3f3f3f3f;
  int n, m;
  int a[MAXN];
  char s[MAXN][MAXM];
  int p[MAXM];
  inline long long get_res(int mask) {
    long long has = 0;
    static int coef[MAXM];
    memset(coef, 0, (m + 1) << 2);
    for (int i = 1; i <= n; ++i) {
      int t = (((mask) >> (i - 1)) & 1) ? 1 : -1;
      has += -t * a[i];
      for (int j = 1; j <= m; ++j)
        if (s[i][j] == '1') coef[j] += t;
    }
    static int id[MAXM];
    for (int i = 1; i <= m; ++i) id[i] = i;
    sort(id + 1, id + m + 1, [](int x, int y) { return coef[x] < coef[y]; });
    for (int i = 1; i <= m; ++i) p[id[i]] = i;
    for (int i = 1; i <= m; ++i) has += coef[i] * p[i];
    return has;
  }
  void solve(void) {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
    for (int i = 1; i <= n; ++i) scanf("%s", s[i] + 1);
    int all = (1 << n) - 1;
    long long ans = -linf;
    int ansmask = -1;
    for (int mask = 0; mask <= all; ++mask) {
      long long cur = get_res(mask);
      if (cur > ans) ans = cur, ansmask = mask;
    }
    get_res(ansmask);
    for (int i = 1; i <= m; ++i) printf("%d ", p[i]);
    printf("\n");
  }
  int main(void) {
    int T;
    scanf("%d", &T);
    while (T--) solve();
    return 0;
  }
  