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;
  const int MAXM = 10000;
  char buff[MAXM + 10];
  int x[MAXN];
  int a[MAXN][MAXM];
  int ac[MAXN];
  int c[MAXM];
  int d[MAXN * 2 + 2];
  int p[MAXM];
  int pi[MAXM];
  int best[MAXM];
  void solve() {
    int n, m;
    scanf("%d%d", &n, &m);
    for (int i = 0; i < n; ++i) scanf("%d", &x[i]);
    fill(ac, ac + n, 0);
    for (int i = 0; i < n; ++i) {
      scanf("\n%s", buff);
      for (int j = 0; j < m; ++j)
        if (buff[j] == '1') {
          a[i][ac[i]] = j;
          ++ac[i];
        }
    }
    long long ans = -(1LL << 62);
    int mask_n = 1 << n;
    for (int mask = 0; mask < mask_n; ++mask) {
      fill(c, c + m, 0);
      for (int i = 0; i < n; ++i) {
        if ((mask >> i) & 1) {
          int z = ac[i];
          for (int j = 0; j < z; ++j) c[a[i][j]] -= 1;
        } else {
          int z = ac[i];
          for (int j = 0; j < z; ++j) c[a[i][j]] += 1;
        }
      }
      fill(d, d + (2 * n + 2), 0);
      for (int j = 0; j < m; ++j) ++d[c[j] + n + 1];
      for (int j = 1; j < 2 * n + 2; ++j) d[j] += d[j - 1];
      for (int j = 0; j < m; ++j) {
        int &z = d[c[j] + n];
        pi[z] = j;
        ++z;
      }
      for (int j = 0; j < m; ++j) p[pi[j]] = j + 1;
      long long res = 0;
      for (int i = 0; i < n; ++i) {
        int y = 0;
        int z = ac[i];
        for (int j = 0; j < z; ++j) y += p[a[i][j]];
        res += abs(y - x[i]);
      }
      if (ans < res) {
        ans = res;
        for (int j = 0; j < m; ++j) best[j] = p[j];
      }
    }
    for (int i = 0; i < m; ++i) printf("%d ", best[i]);
    printf("\n");
  }
  int main() {
    int t;
    scanf("%d", &t);
    for (int i = 0; i < t; ++i) solve();
    return 0;
  }
  