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>
  long long MOD = 1e9 + 7;
  long long bigMOD = 998244353;
  using namespace std;
  struct custom_hash {
    static uint64_t splitmix64(uint64_t x) {
      x += 0x9e3779b97f4a7c15;
      x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
      x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
      return x ^ (x >> 31);
    }
    size_t operator()(uint64_t x) const {
      static const uint64_t FIXED_RANDOM =
          chrono::steady_clock::now().time_since_epoch().count();
      return splitmix64(x + FIXED_RANDOM);
    }
  };
  long long exp2(long long base, int power) {
    if (power == 0) return 1;
    if (power == 1) return (base + MOD) % MOD;
    long long ans = exp2(base, power / 2);
    ans = (ans * ans + MOD) % MOD;
    if (power % 2 == 1) ans = (ans * base + MOD) % MOD;
    return (ans + MOD) % MOD;
  }
  unsigned long long power(unsigned long long x, int y, int p) {
    unsigned long long res = 1;
    x = x % p;
    while (y > 0) {
      if (y & 1) res = (res * x) % p;
      y = y >> 1;
      x = (x * x) % p;
    }
    return res;
  }
  unsigned long long modInverse(unsigned long long n, int p) {
    return power(n, p - 2, p);
  }
  unsigned long long nCr(unsigned long long n, int r, int p) {
    if (n < r) return 0;
    if (r == 0) return 1;
    vector<unsigned long long> fac;
    fac.resize(n + 1);
    fac[0] = 1;
    for (int i = 1; i <= n; i++) fac[i] = (fac[i - 1] * i) % p;
    return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) %
           p;
  }
  struct BIT {
    int n;
    vector<int> B;
    BIT(vector<int> A) {
      n = A.size();
      B.resize(n + 1);
      for (int k = 0; k < n; k++) {
        addValue(k, A[k]);
      }
    }
    int getSum(int m) {
      m++;
      int sum = 0;
      while (m > 0) {
        sum += B[m];
        m -= m & (-m);
      }
      return sum;
    }
    void addValue(int m, int increase) {
      m++;
      while (m <= n) {
        B[m] += increase;
        m += (m) & (-m);
      }
    }
  };
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    int aa;
    cin >> aa;
    for (int kk = 0; kk < aa; kk++) {
      long long maxi = -1e18;
      vector<pair<int, int>> anss;
      int n, m;
      cin >> n >> m;
      vector<long long> ex;
      long long sum = 0;
      for (int i = 0; i < n; i++) {
        int a;
        cin >> a;
        ex.push_back(a);
        sum += a;
      }
      vector<vector<char>> solved;
      for (int i = 0; i < n; i++) {
        solved.push_back(vector<char>());
        for (int j = 0; j < m; j++) {
          char a;
          cin >> a;
          solved[i].push_back(a);
        }
      }
      for (int k = 0; k < pow(2, n); k++) {
        vector<int> dig;
        int temp = k;
        for (int i = 0; i < n; i++) {
          dig.push_back(temp % 2);
          temp /= 2;
        }
        vector<int> perQ;
        perQ.resize(m);
        for (int i = 0; i < n; i++) {
          for (int j = 0; j < m; j++) {
            if (solved[i][j] == '1') {
              if (dig[i] == 0) {
                perQ[j]--;
              } else {
                perQ[j]++;
              }
            }
          }
        }
        vector<pair<int, int>> order;
        for (int i = 0; i < m; i++) {
          order.push_back(make_pair(perQ[i], i));
        }
        sort(order.begin(), order.end());
        long long sum = 0;
        for (int i = 0; i < m; i++) {
          sum += (long long)order[i].first * (i + 1);
        }
        for (int i = 0; i < n; i++) {
          if (dig[i] == 0) {
            sum += ex[i];
          } else {
            sum -= ex[i];
          }
        }
        if (sum > maxi) {
          maxi = sum;
          anss = order;
        }
      }
      vector<int> ans;
      ans.resize(m);
      for (int i = 0; i < m; i++) {
        ans[anss[i].second] = i + 1;
      }
      for (int i = 0; i < m; i++) {
        cout << ans[i] << " ";
      }
      cout << endl;
    }
  }
  