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>
  const int N = 10;
  const int M = 10000;
  int n, m, expected[N], ok[N][M], coef[M], cnt[N + N + 1], rank[M];
  char buffer[M + 1];
  std::pair<int, int> best, order[M];
  void dfs(int i, int mask, int coef0) {
    if (i < n) {
      for (int j = 0; j < m; ++j) {
        if (ok[i][j]) {
          coef[j]--;
        }
      }
      dfs(i + 1, mask, coef0 + expected[i]);
      for (int j = 0; j < m; ++j) {
        if (ok[i][j]) {
          coef[j] += 2;
        }
      }
      dfs(i + 1, mask | 1 << i, coef0 - expected[i]);
      for (int j = 0; j < m; ++j) {
        if (ok[i][j]) {
          coef[j]--;
        }
      }
    } else {
      for (int j = 0; j < m; ++j) {
        cnt[n + coef[j]]++;
      }
      int cur = coef0;
      int rank = 1;
      for (int c = -n; c <= n; ++c) {
        while (cnt[n + c] > 0) {
          cur += c * rank;
          rank++;
          cnt[n + c]--;
        }
      }
      best = std::max(best, {cur, mask});
    }
  }
  int main() {
    int T;
    scanf("%d", &T);
    while (T--) {
      scanf("%d%d", &n, &m);
      for (int i = 0; i < n; ++i) {
        scanf("%d", expected + i);
      }
      for (int i = 0; i < n; ++i) {
        scanf("%s", buffer);
        for (int j = 0; j < m; ++j) {
          ok[i][j] = buffer[j] == '1';
        }
      }
      best = {-1, -1};
      std::fill(coef, coef + m, 0);
      dfs(0, 0, 0);
      int mask = best.second;
      for (int i = 0; i < n; ++i) {
        if (mask >> i & 1) {
          for (int j = 0; j < m; ++j) {
            coef[j] += ok[i][j];
          }
        } else {
          for (int j = 0; j < m; ++j) {
            coef[j] -= ok[i][j];
          }
        }
      }
      for (int j = 0; j < m; ++j) {
        order[j] = {coef[j], j};
      }
      std::sort(order, order + m);
      int cur = 1;
      for (int j = 0; j < m; ++j) {
        rank[order[j].second] = cur++;
      }
      for (int j = 0; j < m; ++j) {
        printf("%d%c", rank[j], " \n"[j + 1 == m]);
      }
    }
  }
  