#include <bits/stdc++.h>
using namespace std;
void testCase() {
  int n, m;
  cin >> n >> m;
  vector<int> x(n);
  for (int &it : x) {
    cin >> it;
  }
  vector<string> res(n);
  for (int i = 0; i < n; ++i) {
    cin >> res[i];
  }
  int64_t best = -1;
  vector<int> sol(m);
  for (int mask = 0; mask < (1 << n); ++mask) {
    vector<pair<int, int>> q(m);
    for (int i = 0; i < m; ++i) {
      q[i].second = i;
    }
    int64_t cost = 0;
    for (int i = 0; i < n; ++i) {
      if (mask & (1 << i)) {
        cost += x[i];
        for (int j = 0; j < m; ++j) {
          if (res[i][j] == '1') {
            q[j].first -= 1;
          }
        }
      } else {
        cost -= x[i];
        for (int j = 0; j < m; ++j) {
          if (res[i][j] == '1') {
            q[j].first += 1;
          }
        }
      }
    }
    sort(q.begin(), q.end());
    for (int i = 0; i < m; ++i) {
      cost += (int64_t)(i + 1) * q[i].first;
    }
    if (best < cost) {
      best = cost;
      for (int i = 0; i < m; ++i) {
        sol[q[i].second] = i;
      }
    }
  }
  for (int i = 0; i < m; ++i) {
    cout << sol[i] + 1 << ' ';
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int tests;
  cin >> tests;
  for (int tc = 0; tc < tests; ++tc) {
    testCase();
  }
  return 0;
}

### 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;
void testCase() {
  int n, m;
  cin >> n >> m;
  vector<int> x(n);
  for (int &it : x) {
    cin >> it;
  }
  vector<string> res(n);
  for (int i = 0; i < n; ++i) {
    cin >> res[i];
  }
  int64_t best = -1;
  vector<int> sol(m);
  for (int mask = 0; mask < (1 << n); ++mask) {
    vector<pair<int, int>> q(m);
    for (int i = 0; i < m; ++i) {
      q[i].second = i;
    }
    int64_t cost = 0;
    for (int i = 0; i < n; ++i) {
      if (mask & (1 << i)) {
        cost += x[i];
        for (int j = 0; j < m; ++j) {
          if (res[i][j] == '1') {
            q[j].first -= 1;
          }
        }
      } else {
        cost -= x[i];
        for (int j = 0; j < m; ++j) {
          if (res[i][j] == '1') {
            q[j].first += 1;
          }
        }
      }
    }
    sort(q.begin(), q.end());
    for (int i = 0; i < m; ++i) {
      cost += (int64_t)(i + 1) * q[i].first;
    }
    if (best < cost) {
      best = cost;
      for (int i = 0; i < m; ++i) {
        sol[q[i].second] = i;
      }
    }
  }
  for (int i = 0; i < m; ++i) {
    cout << sol[i] + 1 << ' ';
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int tests;
  cin >> tests;
  for (int tc = 0; tc < tests; ++tc) {
    testCase();
  }
  return 0;
}

```