#include <bits/stdc++.h>
using namespace std;
char s[10][10001];
int X[10];
vector<pair<int, int>> v;
vector<int> p;
pair<int, vector<int>> Calculate(int n, int m, int mask) {
  for (int k = 0; k < m; ++k) {
    int c = 0;
    for (int i = 0; i < n; ++i) {
      if (s[i][k] == '0') {
        continue;
      }
      c += ((mask & (1 << i)) != 0) ? 1 : -1;
    }
    v[k] = {c, k};
  }
  int ans = 0;
  for (int i = 0; i < n; ++i) {
    int c = ((mask & (1 << i)) != 0) ? -1 : 1;
    ans += c * X[i];
  }
  sort(v.begin(), v.end());
  for (int k = 0; k < m; ++k) {
    ans += v[k].first * (k + 1);
    p[v[k].second] = k + 1;
  }
  return {ans, p};
}
void Solve() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &X[i]);
  }
  for (int i = 0; i < n; ++i) {
    scanf("%s", s[i]);
  }
  pair<int, vector<int>> ans = {INT_MIN, vector<int>()};
  v.resize(m), p.resize(m);
  for (int mask = 0; mask < (1 << n); ++mask) {
    ans = max(ans, Calculate(n, m, mask));
  }
  auto [_, p] = ans;
  for (int pi : p) {
    printf("%d ", pi);
  }
  printf("\n");
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    Solve();
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 s[10][10001];
int X[10];
vector<pair<int, int>> v;
vector<int> p;
pair<int, vector<int>> Calculate(int n, int m, int mask) {
  for (int k = 0; k < m; ++k) {
    int c = 0;
    for (int i = 0; i < n; ++i) {
      if (s[i][k] == '0') {
        continue;
      }
      c += ((mask & (1 << i)) != 0) ? 1 : -1;
    }
    v[k] = {c, k};
  }
  int ans = 0;
  for (int i = 0; i < n; ++i) {
    int c = ((mask & (1 << i)) != 0) ? -1 : 1;
    ans += c * X[i];
  }
  sort(v.begin(), v.end());
  for (int k = 0; k < m; ++k) {
    ans += v[k].first * (k + 1);
    p[v[k].second] = k + 1;
  }
  return {ans, p};
}
void Solve() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &X[i]);
  }
  for (int i = 0; i < n; ++i) {
    scanf("%s", s[i]);
  }
  pair<int, vector<int>> ans = {INT_MIN, vector<int>()};
  v.resize(m), p.resize(m);
  for (int mask = 0; mask < (1 << n); ++mask) {
    ans = max(ans, Calculate(n, m, mask));
  }
  auto [_, p] = ans;
  for (int pi : p) {
    printf("%d ", pi);
  }
  printf("\n");
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    Solve();
  }
  return 0;
}

```