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 
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const long long MAX = 1e6;
void solve() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long n, m;
  cin >> n >> m;
  vector<long long> x(n);
  for (long long i = 0; i < n; i++) cin >> x[i];
  vector<string> s(n);
  for (long long i = 0; i < n; i++) cin >> s[i];
  vector<long long> ans;
  long long mx = -1;
  for (long long mask = 0; mask < (1 << n); mask++) {
    long long cur = 0;
    for (long long i = 0; i < n; i++) {
      if (mask & (1 << i))
        cur += x[i];
      else
        cur -= x[i];
    }
    vector<long long> subjects[2 * n + 1];
    for (long long j = 0; j < m; j++) {
      long long t = 0;
      for (long long i = 0; i < n; i++) {
        if (s[i][j] == '1') {
          if (mask & (1 << i))
            t--;
          else
            t++;
        }
      }
      subjects[t + n].push_back(j);
    }
    vector<long long> present(m);
    long long nxt = 1;
    for (long long i = 0; i < 2 * n + 1; i++) {
      for (auto j : subjects[i]) {
        present[j] = nxt;
        cur += nxt * (i - n);
        nxt++;
      }
    }
    if (cur > mx) {
      mx = cur;
      ans = present;
    }
  }
  for (auto i : ans) cout << i << " ";
  cout << '\n';
  return;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}
