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 comment(linker, "/stack:336777216")
  #pragma GCC optimize("Ofast,no-stack-protector")
  #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
  using namespace std;
  int MSK, N, owo[1 << 10];
  int getoffon(int x) {
    int sum = 0;
    for (int i = 0; i < N; i++) {
      if (x & (1 << i)) {
        if (MSK & (1 << i))
          sum++;
        else
          sum--;
      }
    }
    return sum;
  }
  bool cmp(int a, int b) {
    if (owo[a] == owo[b]) return a < b;
    return owo[a] > owo[b];
  }
  vector<int> pos[1024];
  int cnt[1024];
  void solve() {
    int n, m;
    cin >> n >> m;
    N = n;
    vector<int> x(n);
    for (int i = 0; i < n; i++) {
      cin >> x[i];
    }
    memset(owo, 0, sizeof(owo));
    vector<string> s(n);
    for (int i = 0; i < n; i++) {
      cin >> s[i];
    }
    for (int i = 0; i < m; i++) {
      int msk = 0;
      for (int j = 0; j < n; j++) {
        if (s[j][i] == '1') msk |= (1 << j);
      }
      pos[msk].push_back(i);
      cnt[msk]++;
    }
    vector<int> msks;
    for (int i = 0; i < (1 << n); i++) {
      msks.push_back(i);
    }
    long long ans = -1;
    int best = 0;
    for (int msk = 0; msk < (1 << n); msk++) {
      long long tmp = 0;
      for (int j = 0; j < n; j++) {
        if (msk & (1 << j))
          tmp -= x[j];
        else
          tmp += x[j];
      }
      MSK = msk;
      vector<int> ccnt(21);
      for (int i = 0; i < (1 << n); i++) {
        if (cnt[i] == 0) continue;
        ccnt[getoffon(i) + 10] += cnt[i];
      }
      int petr = 1;
      for (int i = 0; i <= 20; i++) {
        int mul = i - 10;
        int cc = ccnt[i];
        if (cc == 0) continue;
        int l = petr - 1, r = petr + cc - 1;
        long long tp = 1LL * r * (r + 1) / 2;
        tp -= 1LL * l * (l + 1) / 2;
        tmp += 1LL * tp * mul;
        petr = r + 1;
      }
      if (tmp > ans) {
        ans = tmp;
        best = msk;
      }
    }
    vector<int> p(m);
    for (int msk = best; msk <= best; msk++) {
      MSK = msk;
      for (int i = 0; i < (1 << n); i++) {
        owo[i] = getoffon(i);
      }
      sort(msks.begin(), msks.end(), cmp);
      int petr = m;
      for (int i = 0; i < (1 << n); i++) {
        for (int j : pos[msks[i]]) {
          p[j] = petr;
          petr--;
        }
      }
    }
    for (int i : p) cout << i << ' ';
    cout << '\n';
    for (int i = 0; i < m; i++) {
      int msk = 0;
      for (int j = 0; j < n; j++) {
        if (s[j][i] == '1') msk |= (1 << j);
      }
      pos[msk].clear();
      cnt[msk] = 0;
    }
  }
  signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int t;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  