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;
  using LL = long long;
  const int MAXN = 11;
  const int MAXM = 1e4 + 5;
  int n, m, ex[MAXN];
  int cnt[30];
  char w[MAXN][MAXM];
  void solve() {
    scanf("%d %d", &n, &m);
    for (int i = 0; i < n; i++) scanf("%d", &ex[i]);
    for (int i = 0; i < n; i++) scanf("%s", w[i]);
    LL ans = 0;
    int pos;
    for (int i = 0; i < (1 << n); i++) {
      for (int j = 0; j <= 20; j++) cnt[j] = 0;
      for (int j = 0; j < m; j++) {
        int x[2] = {0, 0};
        for (int k = 0; k < n; k++)
          if (w[k][j] == '1') x[((1 << k) & i) != 0]++;
        cnt[x[1] - x[0] + 10]++;
      }
      int mm = m;
      LL sum = 0;
      for (int j = 0; j < n; j++)
        if (((1 << j) & i) == 0)
          sum += ex[j];
        else
          sum -= ex[j];
      for (int j = 20; j >= 0; j--) {
        sum += 1ll * (j - 10) * (2 * mm - cnt[j] + 1) * cnt[j] / 2;
        mm -= cnt[j];
      }
      if (sum >= ans) ans = sum, pos = i;
    }
    {
      for (int j = 0; j <= 20; j++) cnt[j] = 0;
      for (int j = 0; j < m; j++) {
        int x[2] = {0, 0};
        for (int k = 0; k < n; k++)
          if (w[k][j] == '1') x[((1 << k) & pos) != 0]++;
        cnt[x[1] - x[0] + 10]++;
      }
      for (int j = 1; j <= 20; j++) cnt[j] += cnt[j - 1];
      for (int j = 0; j < m; j++) {
        int x[2] = {0, 0};
        for (int k = 0; k < n; k++)
          if (w[k][j] == '1') x[((1 << k) & pos) != 0]++;
        printf("%d ", cnt[x[1] - x[0] + 10]);
        cnt[x[1] - x[0] + 10]--;
      }
      printf("\n");
    }
  }
  int main() {
    int T;
    scanf("%d", &T);
    for (int i = 1; i <= T; i++) {
      solve();
    }
    return 0;
  }
  