#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n, m;
    scanf("%d%d", &n, &m);
    vector<int> x(n);
    for (int i = 0; i < int(n); ++i) scanf("%d", &x[i]);
    vector<vector<int>> a(n, vector<int>(m));
    for (int i = 0; i < int(n); ++i)
      for (int j = 0; j < int(m); ++j) scanf("%1d", &a[i][j]);
    int ans = -1;
    vector<int> best;
    for (int mask = 0; mask < int(1 << n); ++mask) {
      vector<int> val(m);
      for (int i = 0; i < int(n); ++i)
        for (int j = 0; j < int(m); ++j)
          if (a[i][j]) val[j] += ((mask >> i) & 1) ? +1 : -1;
      int res = 0;
      for (int i = 0; i < int(n); ++i) res += ((mask >> i) & 1) ? -x[i] : x[i];
      vector<int> p(m);
      iota(p.begin(), p.end(), 0);
      sort(p.begin(), p.end(), [&](int x, int y) { return val[x] < val[y]; });
      for (int i = 0; i < int(m); ++i) res += val[p[i]] * (i + 1);
      if (res > ans) ans = res, best = p;
    }
    vector<int> ansPerm(m);
    for (int i = 0; i < int(m); ++i) ansPerm[best[i]] = i;
    for (int i = 0; i < int(m); ++i) printf("%d ", ansPerm[i] + 1);
    puts("");
  }
}

### Prompt

In Cpp, your task is to solve 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() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n, m;
    scanf("%d%d", &n, &m);
    vector<int> x(n);
    for (int i = 0; i < int(n); ++i) scanf("%d", &x[i]);
    vector<vector<int>> a(n, vector<int>(m));
    for (int i = 0; i < int(n); ++i)
      for (int j = 0; j < int(m); ++j) scanf("%1d", &a[i][j]);
    int ans = -1;
    vector<int> best;
    for (int mask = 0; mask < int(1 << n); ++mask) {
      vector<int> val(m);
      for (int i = 0; i < int(n); ++i)
        for (int j = 0; j < int(m); ++j)
          if (a[i][j]) val[j] += ((mask >> i) & 1) ? +1 : -1;
      int res = 0;
      for (int i = 0; i < int(n); ++i) res += ((mask >> i) & 1) ? -x[i] : x[i];
      vector<int> p(m);
      iota(p.begin(), p.end(), 0);
      sort(p.begin(), p.end(), [&](int x, int y) { return val[x] < val[y]; });
      for (int i = 0; i < int(m); ++i) res += val[p[i]] * (i + 1);
      if (res > ans) ans = res, best = p;
    }
    vector<int> ansPerm(m);
    for (int i = 0; i < int(m); ++i) ansPerm[best[i]] = i;
    for (int i = 0; i < int(m); ++i) printf("%d ", ansPerm[i] + 1);
    puts("");
  }
}

```