#include <bits/stdc++.h>
using namespace std;
void solve_case();
signed main() {
  cin.tie(0), cout.tie(0)->sync_with_stdio(0);
  int test;
  cin >> test;
  while (test--) {
    solve_case();
  }
}
const int N = 15, M = 1e4 + 5;
int n, m, a[N], delta[M], p[M], res[M];
vector<int> vec[N * 2];
long long ans;
string s[N];
void solve_case() {
  cin >> n >> m;
  ans = -1;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) cin >> s[i];
  for (int mask = 0; mask < 1 << n; mask++) {
    fill(delta, delta + m, 0);
    long long cur = 0;
    for (int i = 0; i < n; i++) {
      int f = mask >> i & 1 ? 1 : -1;
      cur -= f * a[i];
      for (int j = 0; j < m; j++)
        if (s[i][j] == '1') delta[j] += f;
    }
    for (int i = 0; i < m; i++) vec[delta[i] + N].push_back(i);
    for (int i = 0, f = 1; i < N * 2; i++) {
      for (auto v : vec[i]) {
        cur += f * delta[v];
        p[v] = f++;
      }
      vec[i].clear();
    }
    if (cur > ans) {
      ans = cur;
      for (int i = 0; i < m; i++) res[i] = p[i];
    }
  }
  for (int i = 0; i < m; i++) cout << res[i] << ' ';
  cout << '\n';
}

### Prompt

Create a solution in CPP for 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;
void solve_case();
signed main() {
  cin.tie(0), cout.tie(0)->sync_with_stdio(0);
  int test;
  cin >> test;
  while (test--) {
    solve_case();
  }
}
const int N = 15, M = 1e4 + 5;
int n, m, a[N], delta[M], p[M], res[M];
vector<int> vec[N * 2];
long long ans;
string s[N];
void solve_case() {
  cin >> n >> m;
  ans = -1;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) cin >> s[i];
  for (int mask = 0; mask < 1 << n; mask++) {
    fill(delta, delta + m, 0);
    long long cur = 0;
    for (int i = 0; i < n; i++) {
      int f = mask >> i & 1 ? 1 : -1;
      cur -= f * a[i];
      for (int j = 0; j < m; j++)
        if (s[i][j] == '1') delta[j] += f;
    }
    for (int i = 0; i < m; i++) vec[delta[i] + N].push_back(i);
    for (int i = 0, f = 1; i < N * 2; i++) {
      for (auto v : vec[i]) {
        cur += f * delta[v];
        p[v] = f++;
      }
      vec[i].clear();
    }
    if (cur > ans) {
      ans = cur;
      for (int i = 0; i < m; i++) res[i] = p[i];
    }
  }
  for (int i = 0; i < m; i++) cout << res[i] << ' ';
  cout << '\n';
}

```