#include <bits/stdc++.h>
using namespace std;
char buff[100000];
void solve() {
  int n, m;
  scanf("%d%d", &n, &m);
  vector<int> x(n);
  for (int i = 0; i < n; ++i) scanf("%d", &x[i]);
  vector<vector<int>> a(n);
  for (int i = 0; i < n; ++i) {
    scanf("\n%s", buff);
    for (int j = 0; j < m; ++j)
      if (buff[j] == '1') a[i].push_back(j);
  }
  long long ans = -(1LL << 62);
  vector<int> best;
  int mask_n = 1 << n;
  for (int mask = 0; mask < mask_n; ++mask) {
    vector<int> c(m);
    for (int i = 0; i < n; ++i) {
      if ((mask >> i) & 1) {
        for (int j = 0; j < a[i].size(); ++j) c[a[i][j]] -= 1;
      } else {
        for (int j = 0; j < a[i].size(); ++j) c[a[i][j]] += 1;
      }
    }
    vector<pair<int, int>> d(m);
    for (int j = 0; j < m; ++j) d[j] = pair<int, int>(c[j], j);
    sort(d.begin(), d.end());
    vector<int> p(m);
    for (int j = 0; j < m; ++j) p[d[j].second] = j + 1;
    long long res = 0;
    for (int i = 0; i < n; ++i) {
      int y = 0;
      for (int j = 0; j < a[i].size(); ++j) y += p[a[i][j]];
      res += abs(y - x[i]);
    }
    if (ans < res) {
      ans = res;
      best = p;
    }
  }
  for (int i = 0; i < m; ++i) printf("%d ", best[i]);
  printf("\n");
}
int main() {
  int t;
  scanf("%d", &t);
  for (int i = 0; i < t; ++i) solve();
  return 0;
}

### Prompt

Your task is to create 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;
char buff[100000];
void solve() {
  int n, m;
  scanf("%d%d", &n, &m);
  vector<int> x(n);
  for (int i = 0; i < n; ++i) scanf("%d", &x[i]);
  vector<vector<int>> a(n);
  for (int i = 0; i < n; ++i) {
    scanf("\n%s", buff);
    for (int j = 0; j < m; ++j)
      if (buff[j] == '1') a[i].push_back(j);
  }
  long long ans = -(1LL << 62);
  vector<int> best;
  int mask_n = 1 << n;
  for (int mask = 0; mask < mask_n; ++mask) {
    vector<int> c(m);
    for (int i = 0; i < n; ++i) {
      if ((mask >> i) & 1) {
        for (int j = 0; j < a[i].size(); ++j) c[a[i][j]] -= 1;
      } else {
        for (int j = 0; j < a[i].size(); ++j) c[a[i][j]] += 1;
      }
    }
    vector<pair<int, int>> d(m);
    for (int j = 0; j < m; ++j) d[j] = pair<int, int>(c[j], j);
    sort(d.begin(), d.end());
    vector<int> p(m);
    for (int j = 0; j < m; ++j) p[d[j].second] = j + 1;
    long long res = 0;
    for (int i = 0; i < n; ++i) {
      int y = 0;
      for (int j = 0; j < a[i].size(); ++j) y += p[a[i][j]];
      res += abs(y - x[i]);
    }
    if (ans < res) {
      ans = res;
      best = p;
    }
  }
  for (int i = 0; i < m; ++i) printf("%d ", best[i]);
  printf("\n");
}
int main() {
  int t;
  scanf("%d", &t);
  for (int i = 0; i < t; ++i) solve();
  return 0;
}

```