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>
  int ww[10000];
  void sort(int *jj, int m) {
    static int kk[(10 + 1) * 2], jj_[10000];
    int j, w;
    memset(kk, 0, sizeof kk);
    for (j = 0; j < m; j++) {
      w = 10 + ww[j];
      kk[w + 1]++;
    }
    for (w = 1; w < (10 + 1) * 2; w++) kk[w] += kk[w - 1];
    for (j = 0; j < m; j++) {
      w = 10 + ww[j];
      jj_[kk[w]++] = jj[j];
    }
    memcpy(jj, jj_, m * sizeof *jj_);
  }
  int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
      static int xx[10], jj[10000], jj_[10000], pp[10000];
      static char cc[10][10000 + 1];
      int n, m, i, j, b, sum_, sum;
      scanf("%d%d", &n, &m);
      for (i = 0; i < n; i++) scanf("%d", &xx[i]);
      for (i = 0; i < n; i++) scanf("%s", cc[i]);
      sum_ = -1;
      for (b = 0; b < 1 << n; b++) {
        memset(ww, 0, m * sizeof *ww);
        sum = 0;
        for (i = 0; i < n; i++)
          if ((b & 1 << i) == 0) {
            sum += xx[i];
            for (j = 0; j < m; j++)
              if (cc[i][j] == '1') ww[j]--;
          } else {
            sum -= xx[i];
            for (j = 0; j < m; j++)
              if (cc[i][j] == '1') ww[j]++;
          }
        for (j = 0; j < m; j++) jj[j] = j;
        sort(jj, m);
        for (j = 0; j < m; j++) sum += ww[jj[j]] * (j + 1);
        if (sum_ < sum) {
          sum_ = sum;
          memcpy(jj_, jj, m * sizeof *jj);
        }
      }
      for (j = 0; j < m; j++) pp[jj_[j]] = j + 1;
      for (j = 0; j < m; j++) printf("%d ", pp[j]);
      printf("\n");
    }
    return 0;
  }
  