#include <bits/stdc++.h>
using namespace std;
long long a[15];
int k[10005];
char s[11][10005];
vector<long long> v[200005], vv, save;
int outs[20005];
int main() {
  long long ans;
  int t;
  cin >> t;
  while (t--) {
    ans = -1e18;
    int n, m;
    cin >> n >> m;
    for (int i = 0; i < n; i++) scanf("%lld", a + i);
    for (int i = 0; i < n; i++) scanf("%s", s[i]);
    for (int i = 0; i < (1 << n); i++) {
      for (int j = 0; j <= 20; j++) v[j].clear();
      vv.clear();
      long long c = 0;
      for (int j = 0; j < n; j++) {
        if (i & (1 << j))
          c += a[j];
        else
          c -= a[j];
      }
      for (int j = 0; j < m; j++) k[j] = 0;
      for (int j = 0; j < n; j++) {
        if (i & (1 << j)) {
          for (int q = 0; q < m; q++)
            if (s[j][q] == '1') k[q]--;
        } else {
          for (int q = 0; q < m; q++)
            if (s[j][q] == '1') k[q]++;
        }
      }
      for (int j = 0; j < m; j++) {
        v[k[j] + 10].push_back(j);
      }
      for (int j = 0; j <= 20; j++)
        for (auto &k : v[j]) vv.push_back(k);
      for (int j = 0; j < m; j++) {
        c += k[vv[j]] * (j + 1);
      }
      if (c > ans) {
        ans = c;
        save = vv;
      }
    }
    int cnt = 0;
    for (auto k : save) {
      outs[k] = ++cnt;
    }
    for (int i = 0; i < m; i++) printf("%d%c", outs[i], " \n"[i == m - 1]);
  }
  return 0;
}

### 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;
long long a[15];
int k[10005];
char s[11][10005];
vector<long long> v[200005], vv, save;
int outs[20005];
int main() {
  long long ans;
  int t;
  cin >> t;
  while (t--) {
    ans = -1e18;
    int n, m;
    cin >> n >> m;
    for (int i = 0; i < n; i++) scanf("%lld", a + i);
    for (int i = 0; i < n; i++) scanf("%s", s[i]);
    for (int i = 0; i < (1 << n); i++) {
      for (int j = 0; j <= 20; j++) v[j].clear();
      vv.clear();
      long long c = 0;
      for (int j = 0; j < n; j++) {
        if (i & (1 << j))
          c += a[j];
        else
          c -= a[j];
      }
      for (int j = 0; j < m; j++) k[j] = 0;
      for (int j = 0; j < n; j++) {
        if (i & (1 << j)) {
          for (int q = 0; q < m; q++)
            if (s[j][q] == '1') k[q]--;
        } else {
          for (int q = 0; q < m; q++)
            if (s[j][q] == '1') k[q]++;
        }
      }
      for (int j = 0; j < m; j++) {
        v[k[j] + 10].push_back(j);
      }
      for (int j = 0; j <= 20; j++)
        for (auto &k : v[j]) vv.push_back(k);
      for (int j = 0; j < m; j++) {
        c += k[vv[j]] * (j + 1);
      }
      if (c > ans) {
        ans = c;
        save = vv;
      }
    }
    int cnt = 0;
    for (auto k : save) {
      outs[k] = ++cnt;
    }
    for (int i = 0; i < m; i++) printf("%d%c", outs[i], " \n"[i == m - 1]);
  }
  return 0;
}

```