#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    int n, m;
    cin >> n >> m;
    vector<int> a(n), len(n * 2 + 1);
    vector<vector<int> > b(n, vector<int>(m, 0)),
        vec(n * 2 + 1, vector<int>(m, 0));
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        char c;
        cin >> c;
        b[i][j] = c - '0';
      }
    }
    vector<int> res(m);
    int ans = -1;
    for (int i = 0; i < (1 << n); i++) {
      int sum = 0;
      for (int j = 0; j < n; j++) {
        if ((i >> j) & 1)
          sum += a[j];
        else
          sum -= a[j];
      }
      for (int j = 0; j <= n * 2; j++) len[j] = 0;
      for (int j = 0; j < m; j++) {
        int sum2 = n;
        for (int k = 0; k < n; k++) {
          if (b[k][j]) {
            if ((i >> k) & 1)
              --sum2;
            else
              ++sum2;
          }
        }
        vec[sum2][len[sum2]] = j;
        ++len[sum2];
      }
      vector<int> nowres(m);
      int now = 0;
      for (int j = 0; j <= 2 * n; j++) {
        for (int k = 0; k < len[j]; k++) {
          ++now;
          sum += now * (j - n);
          nowres[vec[j][k]] = now;
        }
      }
      if (sum > ans) {
        ans = sum;
        res = nowres;
      }
    }
    for (int i = 0; i < m; i++) cout << res[i] << " ";
    cout << endl;
  }
  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 main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    int n, m;
    cin >> n >> m;
    vector<int> a(n), len(n * 2 + 1);
    vector<vector<int> > b(n, vector<int>(m, 0)),
        vec(n * 2 + 1, vector<int>(m, 0));
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        char c;
        cin >> c;
        b[i][j] = c - '0';
      }
    }
    vector<int> res(m);
    int ans = -1;
    for (int i = 0; i < (1 << n); i++) {
      int sum = 0;
      for (int j = 0; j < n; j++) {
        if ((i >> j) & 1)
          sum += a[j];
        else
          sum -= a[j];
      }
      for (int j = 0; j <= n * 2; j++) len[j] = 0;
      for (int j = 0; j < m; j++) {
        int sum2 = n;
        for (int k = 0; k < n; k++) {
          if (b[k][j]) {
            if ((i >> k) & 1)
              --sum2;
            else
              ++sum2;
          }
        }
        vec[sum2][len[sum2]] = j;
        ++len[sum2];
      }
      vector<int> nowres(m);
      int now = 0;
      for (int j = 0; j <= 2 * n; j++) {
        for (int k = 0; k < len[j]; k++) {
          ++now;
          sum += now * (j - n);
          nowres[vec[j][k]] = now;
        }
      }
      if (sum > ans) {
        ans = sum;
        res = nowres;
      }
    }
    for (int i = 0; i < m; i++) cout << res[i] << " ";
    cout << endl;
  }
  return 0;
}

```