#include <bits/stdc++.h>
using namespace std;
void solve() {
  int n, m;
  cin >> n >> m;
  vector<int> x(n);
  for (int i = 0; i < n; i++) cin >> x[i];
  vector<string> s(n);
  for (int i = 0; i < n; i++) cin >> s[i];
  int64_t best = -1e18;
  vector<int> q(m);
  for (int mask = 0; mask < (1 << n); mask++) {
    int64_t cur = 0;
    for (int i = 0; i < n; i++) {
      if (mask & (1 << i))
        cur += x[i];
      else
        cur -= x[i];
    }
    vector<pair<int, int>> a;
    for (int j = 0; j < m; j++) {
      int coeff = 0;
      for (int i = 0; i < n; i++) {
        if (s[i][j] == '1') {
          if (mask & (1 << i))
            coeff--;
          else
            coeff++;
        }
      }
      a.push_back({coeff, j});
    }
    sort(a.begin(), a.end());
    for (int k = 0; k < m; k++) cur += a[k].first * (k + 1);
    if (cur > best) {
      best = cur;
      for (int k = 0; k < m; k++) {
        int j = a[k].second;
        q[j] = k + 1;
      }
    }
  }
  for (int i = 0; i < m; i++) cout << q[i] << (i + 1 == m ? '\n' : ' ');
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int T;
  cin >> T;
  while (T--) {
    solve();
  }
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
void solve() {
  int n, m;
  cin >> n >> m;
  vector<int> x(n);
  for (int i = 0; i < n; i++) cin >> x[i];
  vector<string> s(n);
  for (int i = 0; i < n; i++) cin >> s[i];
  int64_t best = -1e18;
  vector<int> q(m);
  for (int mask = 0; mask < (1 << n); mask++) {
    int64_t cur = 0;
    for (int i = 0; i < n; i++) {
      if (mask & (1 << i))
        cur += x[i];
      else
        cur -= x[i];
    }
    vector<pair<int, int>> a;
    for (int j = 0; j < m; j++) {
      int coeff = 0;
      for (int i = 0; i < n; i++) {
        if (s[i][j] == '1') {
          if (mask & (1 << i))
            coeff--;
          else
            coeff++;
        }
      }
      a.push_back({coeff, j});
    }
    sort(a.begin(), a.end());
    for (int k = 0; k < m; k++) cur += a[k].first * (k + 1);
    if (cur > best) {
      best = cur;
      for (int k = 0; k < m; k++) {
        int j = a[k].second;
        q[j] = k + 1;
      }
    }
  }
  for (int i = 0; i < m; i++) cout << q[i] << (i + 1 == m ? '\n' : ' ');
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int T;
  cin >> T;
  while (T--) {
    solve();
  }
}

```