#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int tt;
  cin >> tt;
  while (tt--) {
    int n, m;
    cin >> n >> m;
    vector<int> x(n);
    for (int i = 0; i < n; i++) {
      cin >> x[i];
    }
    vector<string> s(n);
    for (int i = 0; i < n; i++) {
      cin >> s[i];
    }
    int ans = -1;
    vector<int> best(m, -1);
    for (int t = 0; t < (1 << n); t++) {
      vector<int> c(m);
      int cur = 0;
      for (int i = 0; i < n; i++) {
        int sign = (t & (1 << i)) ? 1 : -1;
        cur -= x[i] * sign;
        for (int j = 0; j < m; j++) {
          if (s[i][j] == '1') {
            c[j] += sign;
          }
        }
      }
      vector<int> order(m);
      iota(order.begin(), order.end(), 0);
      sort(order.begin(), order.end(),
           [&](int i, int j) { return c[i] < c[j]; });
      for (int i = 0; i < m; i++) {
        cur += c[order[i]] * (i + 1);
      }
      if (cur > ans) {
        ans = cur;
        best = order;
      }
    }
    vector<int> p(m);
    for (int i = 0; i < m; i++) {
      p[best[i]] = i + 1;
    }
    for (int i = 0; i < m; i++) {
      if (i > 0) {
        cout << " ";
      }
      cout << p[i];
    }
    cout << '\n';
  }
  return 0;
}

### Prompt

Generate 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 tt;
  cin >> tt;
  while (tt--) {
    int n, m;
    cin >> n >> m;
    vector<int> x(n);
    for (int i = 0; i < n; i++) {
      cin >> x[i];
    }
    vector<string> s(n);
    for (int i = 0; i < n; i++) {
      cin >> s[i];
    }
    int ans = -1;
    vector<int> best(m, -1);
    for (int t = 0; t < (1 << n); t++) {
      vector<int> c(m);
      int cur = 0;
      for (int i = 0; i < n; i++) {
        int sign = (t & (1 << i)) ? 1 : -1;
        cur -= x[i] * sign;
        for (int j = 0; j < m; j++) {
          if (s[i][j] == '1') {
            c[j] += sign;
          }
        }
      }
      vector<int> order(m);
      iota(order.begin(), order.end(), 0);
      sort(order.begin(), order.end(),
           [&](int i, int j) { return c[i] < c[j]; });
      for (int i = 0; i < m; i++) {
        cur += c[order[i]] * (i + 1);
      }
      if (cur > ans) {
        ans = cur;
        best = order;
      }
    }
    vector<int> p(m);
    for (int i = 0; i < m; i++) {
      p[best[i]] = i + 1;
    }
    for (int i = 0; i < m; i++) {
      if (i > 0) {
        cout << " ";
      }
      cout << p[i];
    }
    cout << '\n';
  }
  return 0;
}

```