#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n, m;
    cin >> n >> m;
    vector<int> a(n);
    for (int i = 0; i < n; ++i) cin >> a[i];
    vector<string> b(n);
    for (int i = 0; i < n; ++i) cin >> b[i];
    vector<int> ans;
    long long mx = -1;
    for (int i = 0; i < 1 << n; ++i) {
      long long surprise = 0;
      vector<int> weight(m);
      for (int j = 0; j < n; ++j) {
        if (i >> j & 1) {
          for (int k = 0; k < m; ++k) {
            if (b[j][k] == '1') weight[k] += 1;
          }
          surprise -= a[j];
        } else {
          for (int k = 0; k < m; ++k) {
            if (b[j][k] == '1') weight[k] -= 1;
          }
          surprise += a[j];
        }
      }
      vector<int> ord(m), r(m);
      iota(ord.begin(), ord.end(), 0);
      sort(ord.begin(), ord.end(),
           [&](int i, int j) { return weight[i] < weight[j]; });
      for (int j = 0; j < m; ++j) {
        r[ord[j]] = j;
        surprise += (j + 1) * weight[ord[j]];
      }
      if (surprise > mx) {
        mx = surprise;
        ans = r;
      }
    }
    for (int i : ans) {
      cout << i + 1 << " ";
    }
    cout << '\n';
  }
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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);
  int t;
  cin >> t;
  while (t--) {
    int n, m;
    cin >> n >> m;
    vector<int> a(n);
    for (int i = 0; i < n; ++i) cin >> a[i];
    vector<string> b(n);
    for (int i = 0; i < n; ++i) cin >> b[i];
    vector<int> ans;
    long long mx = -1;
    for (int i = 0; i < 1 << n; ++i) {
      long long surprise = 0;
      vector<int> weight(m);
      for (int j = 0; j < n; ++j) {
        if (i >> j & 1) {
          for (int k = 0; k < m; ++k) {
            if (b[j][k] == '1') weight[k] += 1;
          }
          surprise -= a[j];
        } else {
          for (int k = 0; k < m; ++k) {
            if (b[j][k] == '1') weight[k] -= 1;
          }
          surprise += a[j];
        }
      }
      vector<int> ord(m), r(m);
      iota(ord.begin(), ord.end(), 0);
      sort(ord.begin(), ord.end(),
           [&](int i, int j) { return weight[i] < weight[j]; });
      for (int j = 0; j < m; ++j) {
        r[ord[j]] = j;
        surprise += (j + 1) * weight[ord[j]];
      }
      if (surprise > mx) {
        mx = surprise;
        ans = r;
      }
    }
    for (int i : ans) {
      cout << i + 1 << " ";
    }
    cout << '\n';
  }
}

```