#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long n, m;
    cin >> n >> m;
    vector<long long> a(n);
    for (int i = 0, ggdem = n; i < ggdem; ++i) cin >> a[i];
    vector<string> s(n);
    for (int i = 0, ggdem = n; i < ggdem; ++i) cin >> s[i];
    long long maxi = 0;
    vector<long long> res(m);
    for (int i = 0, ggdem = (1ll << n); i < ggdem; ++i) {
      vector<pair<long long, long long>> c(m);
      for (int j = 0, ggdem = n; j < ggdem; ++j) {
        long long p = 1;
        if (i & (1ll << j)) p = -1;
        for (int ii = 0, ggdem = m; ii < ggdem; ++ii) {
          if (s[j][ii] == '1') c[ii].first += p;
        }
      }
      for (int ii = 0, ggdem = m; ii < ggdem; ++ii) c[ii].second = ii;
      sort(c.begin(), c.end());
      long long val = 0;
      for (int ii = 0, ggdem = m; ii < ggdem; ++ii)
        val += (ii + 1) * c[ii].first;
      for (int j = 0, ggdem = n; j < ggdem; ++j) {
        if (i & (1ll << j))
          val += a[j];
        else
          val -= a[j];
      }
      if (val >= maxi) {
        maxi = val;
        for (int ii = 0, ggdem = m; ii < ggdem; ++ii) res[c[ii].second] = ii;
      }
    }
    for (auto i : res) cout << i + 1 << " ";
    cout << "\n";
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long n, m;
    cin >> n >> m;
    vector<long long> a(n);
    for (int i = 0, ggdem = n; i < ggdem; ++i) cin >> a[i];
    vector<string> s(n);
    for (int i = 0, ggdem = n; i < ggdem; ++i) cin >> s[i];
    long long maxi = 0;
    vector<long long> res(m);
    for (int i = 0, ggdem = (1ll << n); i < ggdem; ++i) {
      vector<pair<long long, long long>> c(m);
      for (int j = 0, ggdem = n; j < ggdem; ++j) {
        long long p = 1;
        if (i & (1ll << j)) p = -1;
        for (int ii = 0, ggdem = m; ii < ggdem; ++ii) {
          if (s[j][ii] == '1') c[ii].first += p;
        }
      }
      for (int ii = 0, ggdem = m; ii < ggdem; ++ii) c[ii].second = ii;
      sort(c.begin(), c.end());
      long long val = 0;
      for (int ii = 0, ggdem = m; ii < ggdem; ++ii)
        val += (ii + 1) * c[ii].first;
      for (int j = 0, ggdem = n; j < ggdem; ++j) {
        if (i & (1ll << j))
          val += a[j];
        else
          val -= a[j];
      }
      if (val >= maxi) {
        maxi = val;
        for (int ii = 0, ggdem = m; ii < ggdem; ++ii) res[c[ii].second] = ii;
      }
    }
    for (auto i : res) cout << i + 1 << " ";
    cout << "\n";
  }
  return 0;
}

```