#include <bits/stdc++.h>
using namespace std;
using ull = uint64_t;
using ll = int64_t;
using ld = long double;
using pii = pair<int, int>;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int ttt;
  cin >> ttt;
  while (ttt--) {
    int n, m;
    cin >> n >> m;
    vector<int> x(n);
    for (int i = 0; i < n; ++i) cin >> x[i];
    vector<string> v(n);
    for (int i = 0; i < n; ++i) cin >> v[i];
    ll ans = -1e18;
    vector<int> ansp;
    vector<int> coeff;
    vector<int> assignment;
    vector<vector<int>> counter(2 * n + 1);
    for (int mask = 0; mask < (1 << n); ++mask) {
      ll sum = 0;
      coeff.assign(m, 0);
      assignment.assign(m, 0);
      for (int i = 0; i <= 2 * n; ++i) counter[i].clear();
      for (int i = 0; i < n; ++i) {
        bool inv = (mask & (1 << i));
        sum += inv ? x[i] : -x[i];
        for (int j = 0; j < m; ++j) {
          if (v[i][j] == '1') {
            coeff[j] += inv ? -1 : 1;
          }
        }
      }
      for (int j = 0; j < m; ++j) counter[coeff[j] + n].push_back(j);
      int permi = 0;
      for (int i = 0; i <= 2 * n; ++i) {
        for (int j : counter[i]) {
          assignment[j] = ++permi;
          sum += coeff[j] * (ll)permi;
        }
      }
      if (sum > ans) ans = sum, ansp = assignment;
    }
    for (int i : ansp) {
      cout << i << ' ';
    }
    cout << '\n';
  }
}

### Prompt

Please formulate 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;
using ull = uint64_t;
using ll = int64_t;
using ld = long double;
using pii = pair<int, int>;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int ttt;
  cin >> ttt;
  while (ttt--) {
    int n, m;
    cin >> n >> m;
    vector<int> x(n);
    for (int i = 0; i < n; ++i) cin >> x[i];
    vector<string> v(n);
    for (int i = 0; i < n; ++i) cin >> v[i];
    ll ans = -1e18;
    vector<int> ansp;
    vector<int> coeff;
    vector<int> assignment;
    vector<vector<int>> counter(2 * n + 1);
    for (int mask = 0; mask < (1 << n); ++mask) {
      ll sum = 0;
      coeff.assign(m, 0);
      assignment.assign(m, 0);
      for (int i = 0; i <= 2 * n; ++i) counter[i].clear();
      for (int i = 0; i < n; ++i) {
        bool inv = (mask & (1 << i));
        sum += inv ? x[i] : -x[i];
        for (int j = 0; j < m; ++j) {
          if (v[i][j] == '1') {
            coeff[j] += inv ? -1 : 1;
          }
        }
      }
      for (int j = 0; j < m; ++j) counter[coeff[j] + n].push_back(j);
      int permi = 0;
      for (int i = 0; i <= 2 * n; ++i) {
        for (int j : counter[i]) {
          assignment[j] = ++permi;
          sum += coeff[j] * (ll)permi;
        }
      }
      if (sum > ans) ans = sum, ansp = assignment;
    }
    for (int i : ansp) {
      cout << i << ' ';
    }
    cout << '\n';
  }
}

```