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;
  template <class c>
  struct rge {
    c b, e;
  };
  template <class c>
  rge<c> range(c i, c j) {
    return rge<c>{i, j};
  }
  template <class c>
  auto dud(c *x) -> decltype(cerr << *x, 0);
  template <class c>
  char dud(...);
  struct debug {
    template <class c>
    debug &operator<<(const c &) {
      return *this;
    }
  };
  long long __pow(long long x, long long y, long long M) {
    long long res = 1;
    while (y > 0) {
      if (y & 1) res = (res * x) % M;
      y >>= 1;
      x = (x * x) % M;
    }
    return res % M;
  }
  long long __pow(long long x, long long y) {
    long long res = 1;
    while (y > 0) {
      if (y & 1) res = (res * x);
      y >>= 1;
      x = (x * x);
    }
    return res;
  }
  long long mi(long long x, long long M) { return __pow(x, M - 2, M); }
  long long mul(vector<long long> a, long long M) {
    long long res = 1;
    for (auto x : a) res = (res * x) % M;
    return res;
  }
  long long cl(long long a, long long b) {
    if (a % b == 0) return a / b;
    return a / b + 1;
  }
  void umax(long long &a, long long b) { a = max(a, b); }
  void umin(long long &a, long long b) { a = min(a, b); }
  const long long N = 10 + 2, inf = 1e18, M = 1e9 + 7, LN = 20;
  long long Test, n, m, k, q, a[N];
  bool ok = 1;
  string s[N];
  void test_case() {
    cin >> n >> m;
    for (long long i = 0; i < n; i++) cin >> a[i];
    for (long long i = 0; i < n; i++) cin >> s[i];
    long long mx = -inf;
    vector<long long> mxv(m);
    for (long long b = 0; b < (1ll << n); b++) {
      long long res = 0;
      for (long long i = 0; i < n; i++) {
        if (b & (1ll << i))
          res -= a[i];
        else
          res += a[i];
      }
      vector<pair<long long, long long>> t(m);
      for (long long j = 0; j < m; j++) {
        t[j].second = j;
        for (long long i = 0; i < n; i++) {
          if (b & (1ll << i))
            t[j].first += (s[i][j] - '0');
          else
            t[j].first -= (s[i][j] - '0');
        }
      }
      sort((t).begin(), (t).end());
      for (long long i = 0; i < m; i++) res += t[i].first * (i + 1);
      if (mx < res) {
        mx = res;
        for (long long i = 0; i < m; i++) mxv[t[i].second] = i + 1;
      }
    }
    debug() << " ["
            << "mx"
               ": "
            << (mx) << "] ";
    for (auto x : mxv) cout << x << " ";
    cout << '\n';
  }
  int32_t main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    Test = 1;
    cin >> Test;
    for (long long t = 1; t <= Test; t++) {
      test_case();
    }
    return 0;
  }
  