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;
  int T, n, m;
  int x[15];
  char s[15][10005];
  int v[10005], cnt[25], tmp[25];
  int P[10005];
  int ans;
  void dfs(int p, int sum) {
    if (p == n + 1) {
      int cur = m, res = sum;
      for (int i = 20; i >= 0; i--) {
        res += (i - 10) * (cur * 2 - cnt[i] + 1) * cnt[i] / 2;
        cur -= cnt[i];
      }
      tmp[0] = cnt[0];
      for (int i = 1; i <= 20; i++) tmp[i] = tmp[i - 1] + cnt[i];
      if (ans <= res) {
        ans = res;
        for (int i = 1; i <= m; i++) {
          P[i] = tmp[v[i] + 10]--;
        }
      }
      return;
    }
    for (int i = 1; i <= m; i++) {
      if (s[p][i] == '1') {
        cnt[v[i] + 10]--;
        v[i]++;
        cnt[v[i] + 10]++;
      }
    }
    dfs(p + 1, sum - x[p]);
    for (int i = 1; i <= m; i++) {
      if (s[p][i] == '1') {
        cnt[v[i] + 10]--;
        v[i] -= 2;
        cnt[v[i] + 10]++;
      }
    }
    dfs(p + 1, sum + x[p]);
    for (int i = 1; i <= m; i++) {
      if (s[p][i] == '1') {
        cnt[v[i] + 10]--;
        v[i]++;
        cnt[v[i] + 10]++;
      }
    }
  }
  void solve() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) scanf("%d", &x[i]);
    for (int i = 1; i <= n; i++) {
      scanf("%s", &s[i][1]);
    }
    for (int i = 1; i <= m; i++) v[i] = 0;
    for (int i = 0; i <= 20; i++) cnt[i] = 0;
    cnt[10] = m;
    ans = 0;
    dfs(1, 0);
    for (int i = 1; i <= m; i++) printf("%d ", P[i]);
    printf("\n");
  }
  int main() {
    scanf("%d", &T);
    while (T--) solve();
    return 0;
  }
  