#include <bits/stdc++.h>
using namespace std;
int e[11];
vector<int> score(10010);
vector<int> ans;
string s[11];
void solve() {
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> e[i];
  for (int i = 1; i <= n; i++) cin >> s[i];
  int up = 1 << n;
  int mx = -1;
  vector<pair<int, int> > ev(m + 1);
  for (int i = 0; i < up; i++) {
    for (int j = 1; j <= m; j++) {
      ev[j].first = 0;
      ev[j].second = j;
    }
    for (int j = 1; j <= n; j++)
      for (int k = 1; k <= m; k++) {
        if (s[j][k - 1] == '1') ev[k].first += (i & 1 << j - 1) ? 1 : -1;
      }
    int sum = 0;
    for (int j = 1; j <= n; j++) sum += (i & 1 << j - 1) ? -e[j] : e[j];
    sort(ev.begin() + 1, ev.end());
    for (int j = 1; j <= m; j++) {
      score[ev[j].second] = j;
      sum += j * ev[j].first;
    }
    if (mx < sum) {
      mx = sum;
      ans = score;
    }
  }
  for (int i = 1; i <= m; i++) cout << ans[i] << " \n"[i == m];
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}

### 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 e[11];
vector<int> score(10010);
vector<int> ans;
string s[11];
void solve() {
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> e[i];
  for (int i = 1; i <= n; i++) cin >> s[i];
  int up = 1 << n;
  int mx = -1;
  vector<pair<int, int> > ev(m + 1);
  for (int i = 0; i < up; i++) {
    for (int j = 1; j <= m; j++) {
      ev[j].first = 0;
      ev[j].second = j;
    }
    for (int j = 1; j <= n; j++)
      for (int k = 1; k <= m; k++) {
        if (s[j][k - 1] == '1') ev[k].first += (i & 1 << j - 1) ? 1 : -1;
      }
    int sum = 0;
    for (int j = 1; j <= n; j++) sum += (i & 1 << j - 1) ? -e[j] : e[j];
    sort(ev.begin() + 1, ev.end());
    for (int j = 1; j <= m; j++) {
      score[ev[j].second] = j;
      sum += j * ev[j].first;
    }
    if (mx < sum) {
      mx = sum;
      ans = score;
    }
  }
  for (int i = 1; i <= m; i++) cout << ans[i] << " \n"[i == m];
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}

```