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 N = 1e4 + 5;
  int t, n, m, cnt[21], q[21][N], r[11], zf[11], res, ans[N], tem, num[N];
  char s[11][N];
  void cal() {
    tem = 0;
    for (int i = 1; i <= n; ++i) tem -= zf[i] * r[i];
    for (int i = 1; i <= m; ++i) num[i] = 0;
    for (int i = 0; i <= 20; ++i) cnt[i] = 0;
    for (int i = 1; i <= n; ++i) {
      for (int j = 1; j <= m; ++j) {
        num[j] += zf[i] * (s[i][j] - '0');
      }
    }
    for (int i = 1; i <= m; ++i) q[num[i] + 10][++cnt[num[i] + 10]] = i;
    int sum = 0;
    for (int i = 0; i <= 20; ++i) {
      tem += (i - 10) * (sum + 1 + sum + cnt[i]) * cnt[i] / 2;
      sum += cnt[i];
    }
  }
  void dfs(int x) {
    if (x == n + 1) {
      cal();
      if (tem > res) {
        res = tem;
        int top = 0;
        for (int i = 0; i <= 20; ++i) {
          for (int j = 1; j <= cnt[i]; ++j) {
            ans[q[i][j]] = ++top;
          }
        }
      }
      return;
    }
    zf[x] = 1;
    dfs(x + 1);
    zf[x] = -1;
    dfs(x + 1);
  }
  void slv() {
    res = -1;
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; ++i) scanf("%d", r + i);
    for (int i = 1; i <= n; ++i) scanf("%s", s[i] + 1);
    dfs(1);
    for (int i = 1; i <= m; ++i) printf("%d ", ans[i]);
    puts("");
  }
  int main() {
    scanf("%d", &t);
    while (t--) slv();
  }
  