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 = 1e4 + 9;
  const long long MOD = 1e9 + 7;
  const int mod = 998244353;
  inline long long qpow(long long b, long long e, long long m) {
    long long a = 1;
    for (; e; e >>= 1, b = b * b % m)
      if (e & 1) a = a * b % m;
    return a;
  }
  long long exgcd(long long a, long long b, long long &x, long long &y) {
    if (b == 0) {
      x = 1, y = 0;
      return a;
    }
    long long d = exgcd(b, a % b, x, y);
    long long z = x;
    x = y, y = z - y * (a / b);
    return d;
  }
  int x[19];
  char s[19][maxn];
  int ans[maxn];
  pair<int, int> val[maxn];
  bool cmp(pair<int, int> A, pair<int, int> B) { return A.second < B.second; }
  int main() {
    int _;
    scanf("%d", &_);
    while (_--) {
      int n, m;
      scanf("%d%d", &n, &m);
      for (int i = 0; i < n; i++) scanf("%d", &x[i]);
      for (int i = 0; i < n; i++) {
        scanf("%s", s[i]);
      }
      int v = -1;
      for (int i = 0; i < (1 << n); i++) {
        int res = 0;
        for (int j = 0; j < n; j++) {
          if (i >> j & 1)
            res += x[j];
          else
            res -= x[j];
        }
        for (int k = 0; k < m; k++) {
          val[k].first = 0;
          val[k].second = k;
        }
        for (int j = 0; j < n; j++) {
          for (int k = 0; k < m; k++) {
            if (s[j][k] == '1') {
              if (i >> j & 1)
                val[k].first--;
              else
                val[k].first++;
            }
          }
        }
        sort(val, val + m);
        for (int j = 0; j < m; j++) {
          res += (j + 1) * val[j].first;
        }
        if (res > v) {
          v = res;
          for (int j = 0; j < m; j++) {
            ans[val[j].second] = j + 1;
          }
        }
      }
      for (int i = 0; i < m; i++) {
        printf("%d ", ans[i]);
      }
      printf("\n");
    }
  }
  