#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n, m;
    cin >> n >> m;
    int x[n];
    for (int i = 0; i < n; i++) cin >> x[i];
    string s[n];
    for (int i = 0; i < n; i++) cin >> s[i];
    int ans[m], score = -1;
    for (int i = 0; i < (1 << n); i++) {
      vector<pair<int, int> > vt;
      for (int j = 0; j < m; j++) {
        int cnt = 0;
        for (int k = 0; k < n; k++) {
          if (s[k][j] == '1') {
            if (i & (1 << k))
              cnt++;
            else
              cnt--;
          }
        }
        vt.push_back(make_pair(cnt, j));
      }
      sort(vt.begin(), vt.end());
      int temp[m], curr_score = 0;
      for (int j = 0; j < m; j++) {
        int idx = vt[j].second;
        temp[idx] = j + 1;
      }
      for (int j = 0; j < n; j++) {
        int sum = 0;
        for (int k = 0; k < m; k++) sum += (s[j][k] - '0') * temp[k];
        curr_score += abs(sum - x[j]);
      }
      if (curr_score > score) {
        score = curr_score;
        for (int j = 0; j < m; j++) ans[j] = temp[j];
      }
    }
    for (int i = 0; i < m; i++) cout << ans[i] << ' ';
    cout << '\n';
  }
}

### Prompt

Your task is to create a CPP solution to the following problem:

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;
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n, m;
    cin >> n >> m;
    int x[n];
    for (int i = 0; i < n; i++) cin >> x[i];
    string s[n];
    for (int i = 0; i < n; i++) cin >> s[i];
    int ans[m], score = -1;
    for (int i = 0; i < (1 << n); i++) {
      vector<pair<int, int> > vt;
      for (int j = 0; j < m; j++) {
        int cnt = 0;
        for (int k = 0; k < n; k++) {
          if (s[k][j] == '1') {
            if (i & (1 << k))
              cnt++;
            else
              cnt--;
          }
        }
        vt.push_back(make_pair(cnt, j));
      }
      sort(vt.begin(), vt.end());
      int temp[m], curr_score = 0;
      for (int j = 0; j < m; j++) {
        int idx = vt[j].second;
        temp[idx] = j + 1;
      }
      for (int j = 0; j < n; j++) {
        int sum = 0;
        for (int k = 0; k < m; k++) sum += (s[j][k] - '0') * temp[k];
        curr_score += abs(sum - x[j]);
      }
      if (curr_score > score) {
        score = curr_score;
        for (int j = 0; j < m; j++) ans[j] = temp[j];
      }
    }
    for (int i = 0; i < m; i++) cout << ans[i] << ' ';
    cout << '\n';
  }
}

```