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>
  #pragma GCC optimize("O4,unroll-loops,no-stack-protector")
  #pragma GCC optimize("Ofast")
  #pragma GCC target("avx,avx2,fma")
  using namespace std;
  int n, m;
  int arr[13][10005];
  int e[13];
  vector<int> ans(10005), Ans(10005);
  signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while (t--) {
      cin >> n >> m;
      for (int i = 0; i < n; i++) cin >> e[i];
      for (int i = 0; i < n; i++) {
        for (int j = 1; j <= m; j++) {
          char c;
          cin >> c;
          arr[i][j] = (c == '1');
        }
      }
      int best = -1;
      for (int i = 0; i <= m; i++) {
        ans[i] = 0;
        Ans[i] = 0;
      }
      for (int bit = 0; bit < (1 << n); bit++) {
        vector<pair<long long, long long> > cnt1(m + 2), cnt2(m + 2);
        cnt1[0] = make_pair(-1, -1);
        cnt2[0] = make_pair(-1, -1);
        for (int i = 1; i <= m; i++) {
          cnt1[i].second = i;
          cnt2[i].second = i;
        }
        for (int i = 0; i < n; i++) {
          for (int j = 1; j <= m; j++)
            if (arr[i][j]) {
              if ((1 << i) & bit)
                cnt1[j].first++;
              else
                cnt1[j].first--;
            }
        }
        sort(cnt1.begin() + 1, cnt1.begin() + 1 + m);
        for (int j = 1; j <= m; j++) {
          ans[cnt1[j].second] = j;
        }
        int res = 0;
        for (int i = 0; i < n; i++) {
          int now = 0;
          for (int j = 1; j <= m; j++)
            if (arr[i][j]) now += ans[j];
          res += abs(now - e[i]);
        }
        if (res > best) {
          best = res;
          Ans = ans;
        }
      }
      for (int i = 1; i <= m; i++) cout << Ans[i] << " ";
      cout << "\n";
    }
  }
  