#include <bits/stdc++.h>
using namespace std;
int x[10], re[10001], anss[10001];
string s[10];
pair<int, int> h[10001];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int T;
  cin >> T;
  while (T--) {
    int n, m;
    cin >> n >> m;
    for (int i = 0; i < n; i++) cin >> x[i];
    for (int i = 0; i < n; i++) cin >> s[i];
    vector<vector<int>> b(n);
    for (int i = 0; i < n; i++)
      for (int j = 1; j <= m; j++)
        if (s[i][j - 1] == '1') b[i].push_back(j);
    int ans = INT_MIN;
    for (int bm = 0; bm < (1 << n); bm++) {
      for (int i = 1; i <= m; i++) h[i] = {0, i};
      for (int i = 0; i < n; i++) {
        int o = (bm >> i) & 1 ? 1 : -1;
        for (int &j : b[i]) h[j].first += o;
      }
      int t = 0;
      for (int i = 0; i < n; i++) t += ((bm >> i) & 1 ? -x[i] : x[i]);
      sort(h + 1, h + m + 1);
      for (int i = 1; i <= m; i++) t += h[i].first * i;
      if (t > ans) {
        ans = t;
        for (int i = 1; i <= m; i++) anss[h[i].second] = i;
      }
    }
    for (int i = 1; i <= m; i++) cout << anss[i] << ' ';
    cout << '\n';
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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 x[10], re[10001], anss[10001];
string s[10];
pair<int, int> h[10001];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int T;
  cin >> T;
  while (T--) {
    int n, m;
    cin >> n >> m;
    for (int i = 0; i < n; i++) cin >> x[i];
    for (int i = 0; i < n; i++) cin >> s[i];
    vector<vector<int>> b(n);
    for (int i = 0; i < n; i++)
      for (int j = 1; j <= m; j++)
        if (s[i][j - 1] == '1') b[i].push_back(j);
    int ans = INT_MIN;
    for (int bm = 0; bm < (1 << n); bm++) {
      for (int i = 1; i <= m; i++) h[i] = {0, i};
      for (int i = 0; i < n; i++) {
        int o = (bm >> i) & 1 ? 1 : -1;
        for (int &j : b[i]) h[j].first += o;
      }
      int t = 0;
      for (int i = 0; i < n; i++) t += ((bm >> i) & 1 ? -x[i] : x[i]);
      sort(h + 1, h + m + 1);
      for (int i = 1; i <= m; i++) t += h[i].first * i;
      if (t > ans) {
        ans = t;
        for (int i = 1; i <= m; i++) anss[h[i].second] = i;
      }
    }
    for (int i = 1; i <= m; i++) cout << anss[i] << ' ';
    cout << '\n';
  }
  return 0;
}

```