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 maxn = 10 + 10, maxm = 1e4 + 10;
  int a[maxn], mat[maxn][maxm], tmp[maxm], cnt[2 * maxn];
  int n, m;
  long long q;
  bool ok;
  long long f(int x) {
    long long res = 0;
    for (int i = 0; i < n; i++) {
      bool c = (1 << i) & x;
      if (c)
        res -= a[i];
      else
        res += a[i];
      for (int j = 1; j <= m; j++) {
        if (mat[i][j]) {
          if (c) {
            tmp[j]++;
          } else {
            tmp[j]--;
          }
        }
      }
    }
    for (int i = 1; i <= m; i++) {
      cnt[n + tmp[i]]++;
    }
    int w = m;
    for (int i = 2 * n; i >= 0; i--) {
      for (int j = 0; j < cnt[i]; j++) {
        res += 1LL * w * (i - n);
        w--;
      }
    }
    if (res == q) {
      ok = true;
      vector<pair<int, int>> v;
      for (int i = 1; i <= m; i++) {
        v.push_back({tmp[i], i});
      }
      sort(v.begin(), v.end());
      for (int i = 0; i < m; i++) {
        tmp[v[i].second] = i + 1;
      }
      for (int i = 1; i <= m; i++) cout << tmp[i] << " ";
      cout << "\n";
    }
    memset(tmp, 0, sizeof(tmp));
    memset(cnt, 0, sizeof(cnt));
    return res;
  }
  int main() {
    ios_base::sync_with_stdio(false), cin.tie(0);
    int t;
    cin >> t;
    while (t--) {
      cin >> n >> m;
      for (int i = 0; i < n; i++) cin >> a[i];
      for (int i = 0; i < n; i++) {
        for (int j = 1; j <= m; j++) {
          char c;
          cin >> c;
          if (c == '1') mat[i][j] = 1;
        }
      }
      long long ans = -1e18;
      ok = false;
      q = -1e18;
      for (int i = 0; i < (1 << n); i++) {
        ans = max(ans, f(i));
      }
      q = ans;
      for (int i = 0; i < (1 << n); i++) {
        f(i);
        if (ok) break;
      }
      for (int i = 0; i < n; i++) {
        for (int j = 1; j <= m; j++) {
          mat[i][j] = 0;
        }
      }
    }
    return 0;
  }
  