#include <bits/stdc++.h>
using namespace std;
const int N = 10, M = 1e4 + 10;
int n, m, first[N], s[M][N], coef[M], ord[M], res[M];
void solve() {
  cin >> n >> m;
  for (int i = 0; i < n; i++) cin >> first[i];
  for (int i = 0; i < n; i++) {
    string b;
    cin >> b;
    for (int j = 0; j < m; j++) s[j + 1][i] = b[j] - '0';
  }
  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 += first[j];
      else
        sum -= first[j];
    }
    for (int j = 1; j <= m; j++) {
      coef[j] = 0;
      for (int k = 0; k < n; k++) {
        if (s[j][k]) coef[j] += (i >> k & 1) ? -1 : 1;
      }
      ord[j] = j;
    }
    sort(ord + 1, ord + m + 1,
         [&](int first, int second) { return coef[first] < coef[second]; });
    for (int j = 1; j <= m; j++) sum += j * coef[ord[j]];
    if (ans < sum) {
      for (int j = 1; j <= m; j++) res[ord[j]] = j;
      ans = sum;
    }
  }
  for (int i = 1; i <= m; i++) cout << res[i] << " ";
  cout << endl;
  return;
}
int main() {
  int t;
  for (cin >> t; t--; solve())
    ;
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 = 10, M = 1e4 + 10;
int n, m, first[N], s[M][N], coef[M], ord[M], res[M];
void solve() {
  cin >> n >> m;
  for (int i = 0; i < n; i++) cin >> first[i];
  for (int i = 0; i < n; i++) {
    string b;
    cin >> b;
    for (int j = 0; j < m; j++) s[j + 1][i] = b[j] - '0';
  }
  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 += first[j];
      else
        sum -= first[j];
    }
    for (int j = 1; j <= m; j++) {
      coef[j] = 0;
      for (int k = 0; k < n; k++) {
        if (s[j][k]) coef[j] += (i >> k & 1) ? -1 : 1;
      }
      ord[j] = j;
    }
    sort(ord + 1, ord + m + 1,
         [&](int first, int second) { return coef[first] < coef[second]; });
    for (int j = 1; j <= m; j++) sum += j * coef[ord[j]];
    if (ans < sum) {
      for (int j = 1; j <= m; j++) res[ord[j]] = j;
      ans = sum;
    }
  }
  for (int i = 1; i <= m; i++) cout << res[i] << " ";
  cout << endl;
  return;
}
int main() {
  int t;
  for (cin >> t; t--; solve())
    ;
  return 0;
}

```