#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
const int M = 35;
int x[M];
int a[M][N];
struct ans {
  int A, B;
} cnt[N];
bool cmp(ans C, ans D) {
  if (C.A == D.A)
    return C.B < D.B;
  else
    return C.A < D.A;
}
int res[N];
void run() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++) {
    scanf("%d", &x[i]);
  }
  for (int i = 0; i < m; i++) cnt[i].A = 0, cnt[i].B = i + 1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      char c;
      cin >> c;
      if (c == '1')
        a[i][j] = 1;
      else
        a[i][j] = 0;
    }
  }
  int nm = (1 << n);
  long long ans = -1e18;
  for (int i = 0; i < nm; i++) {
    long long tot = 0;
    for (int j = 0; j < m; j++) cnt[j].A = 0, cnt[j].B = j + 1;
    for (int j = 0; j < n; j++) {
      if (i >> j & 1)
        tot += x[j];
      else
        tot -= x[j];
    }
    for (int j = 0; j < n; j++) {
      for (int k = 0; k < m; k++) {
        if ((i >> j & 1) && a[j][k])
          cnt[k].A--;
        else if (!(i >> j & 1) && a[j][k])
          cnt[k].A++;
      }
    }
    sort(cnt, cnt + m, cmp);
    for (int j = 0; j < m; j++) tot += cnt[j].A * (j + 1);
    if (tot > ans) {
      ans = tot;
      for (int j = 0; j < m; j++) {
        res[cnt[j].B] = j + 1;
      }
    }
  }
  for (int i = 1; i <= m; i++) {
    printf("%d ", res[i]);
  }
  puts("");
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    run();
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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;
const int N = 1e5 + 5;
const int M = 35;
int x[M];
int a[M][N];
struct ans {
  int A, B;
} cnt[N];
bool cmp(ans C, ans D) {
  if (C.A == D.A)
    return C.B < D.B;
  else
    return C.A < D.A;
}
int res[N];
void run() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++) {
    scanf("%d", &x[i]);
  }
  for (int i = 0; i < m; i++) cnt[i].A = 0, cnt[i].B = i + 1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      char c;
      cin >> c;
      if (c == '1')
        a[i][j] = 1;
      else
        a[i][j] = 0;
    }
  }
  int nm = (1 << n);
  long long ans = -1e18;
  for (int i = 0; i < nm; i++) {
    long long tot = 0;
    for (int j = 0; j < m; j++) cnt[j].A = 0, cnt[j].B = j + 1;
    for (int j = 0; j < n; j++) {
      if (i >> j & 1)
        tot += x[j];
      else
        tot -= x[j];
    }
    for (int j = 0; j < n; j++) {
      for (int k = 0; k < m; k++) {
        if ((i >> j & 1) && a[j][k])
          cnt[k].A--;
        else if (!(i >> j & 1) && a[j][k])
          cnt[k].A++;
      }
    }
    sort(cnt, cnt + m, cmp);
    for (int j = 0; j < m; j++) tot += cnt[j].A * (j + 1);
    if (tot > ans) {
      ans = tot;
      for (int j = 0; j < m; j++) {
        res[cnt[j].B] = j + 1;
      }
    }
  }
  for (int i = 1; i <= m; i++) {
    printf("%d ", res[i]);
  }
  puts("");
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    run();
  }
  return 0;
}

```