#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 10, M = 2 * N, mod = 998244353;
int first[N];
string s[N];
int ans[N];
void solve() {
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> first[i];
  for (int i = 1; i <= n; i++) cin >> s[i], s[i] = " " + s[i];
  long long res = -1, tmp;
  for (int k = 0; k < (1 << n); k++) {
    tmp = 0;
    for (int i = 0; i < n; i++) {
      int r = 1;
      if (k >> i & 1) r = -1;
      tmp += r * first[i + 1];
    }
    vector<pair<int, int> > num;
    for (int i = 1; i <= m; i++) {
      int tmp = 0;
      for (int j = 0; j < n; j++) {
        int r = -1;
        if (k >> j & 1) r = 1;
        tmp += (s[j + 1][i] - '0') * r;
      }
      num.push_back({tmp, i});
    }
    sort((num).begin(), (num).end());
    for (int i = 1; i <= m; i++) tmp += num[i - 1].first * i;
    if (res < tmp) {
      res = tmp;
      for (int i = 0; i < m; i++) {
        auto &[val, id] = num[i];
        ans[id] = i + 1;
      }
    }
  }
  for (int i = 1; i <= m; i++) cout << ans[i] << ' ';
  cout << '\n';
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cout.precision(2), cout.setf(ios::fixed);
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}

### Prompt

Your challenge is to write 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;
const int N = 3e5 + 10, M = 2 * N, mod = 998244353;
int first[N];
string s[N];
int ans[N];
void solve() {
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> first[i];
  for (int i = 1; i <= n; i++) cin >> s[i], s[i] = " " + s[i];
  long long res = -1, tmp;
  for (int k = 0; k < (1 << n); k++) {
    tmp = 0;
    for (int i = 0; i < n; i++) {
      int r = 1;
      if (k >> i & 1) r = -1;
      tmp += r * first[i + 1];
    }
    vector<pair<int, int> > num;
    for (int i = 1; i <= m; i++) {
      int tmp = 0;
      for (int j = 0; j < n; j++) {
        int r = -1;
        if (k >> j & 1) r = 1;
        tmp += (s[j + 1][i] - '0') * r;
      }
      num.push_back({tmp, i});
    }
    sort((num).begin(), (num).end());
    for (int i = 1; i <= m; i++) tmp += num[i - 1].first * i;
    if (res < tmp) {
      res = tmp;
      for (int i = 0; i < m; i++) {
        auto &[val, id] = num[i];
        ans[id] = i + 1;
      }
    }
  }
  for (int i = 1; i <= m; i++) cout << ans[i] << ' ';
  cout << '\n';
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cout.precision(2), cout.setf(ios::fixed);
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}

```