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 <typename T>
  bool chkMax(T &x, T y) {
    return (y > x) ? x = y, 1 : 0;
  }
  template <typename T>
  bool chkMin(T &x, T y) {
    return (y < x) ? x = y, 1 : 0;
  }
  template <typename T>
  void inline read(T &x) {
    int f = 1;
    x = 0;
    char s = getchar();
    while (s < '0' || s > '9') {
      if (s == '-') f = -1;
      s = getchar();
    }
    while (s <= '9' && s >= '0') x = x * 10 + (s ^ 48), s = getchar();
    x *= f;
  }
  const int N = 10, M = 10000;
  int n, m, ans[M];
  pair<int, int> c[M];
  long long x[N], val;
  int s[N][M];
  int main() {
    int T;
    read(T);
    while (T--) {
      val = -9e18;
      read(n), read(m);
      for (int i = 0; i < n; i++) read(x[i]);
      for (int i = 0; i < n; i++)
        for (int j = 0; j < m; j++) scanf("%1d", &s[i][j]);
      for (int i = 0; i < (1 << n); i++) {
        long long ret = 0;
        for (int j = 0; j < m; j++) c[j].first = 0, c[j].second = j;
        for (int j = 0; j < n; j++) {
          int t = (i >> j & 1) ? 1 : -1;
          for (int k = 0; k < m; k++) {
            if (s[j][k]) c[k].first += t;
          }
          if (t == 1)
            ret -= x[j];
          else
            ret += x[j];
        }
        sort(c, c + m);
        for (int j = 0; j < m; j++) ret += c[j].first * (j + 1ll);
        if (chkMax(val, ret)) {
          for (int j = 0; j < m; j++) ans[c[j].second] = j;
        }
      }
      for (int i = 0; i < m; i++) printf("%d ", ans[i] + 1);
      puts("");
    }
    return 0;
  }
  