#include <bits/stdc++.h>
using namespace std;
int n, m;
int a[10], b[10][10005];
void Solve();
int main() {
  int tests = 1;
  scanf("%d", &tests);
  while (tests--) {
    Solve();
  }
  return 0;
}
void Solve() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < (n); ++i) scanf("%d", a + i);
  for (int i = 0; i < (n); ++i)
    for (int j = 1; j <= (m); ++j) scanf("%1d", &b[i][j]);
  static int ans[10005];
  int maxdif = -1;
  for (int s = 0; s < (1 << n); ++s) {
    static int c[10005];
    for (int j = 1; j <= (m); ++j) c[j] = 0;
    for (int i = 0; i < (n); ++i)
      for (int j = 1; j <= (m); ++j) c[j] += b[i][j] ? s >> i & 1 ? -1 : 1 : 0;
    static int per[10005], val[10005];
    for (int j = 1; j <= (m); ++j) per[j] = j;
    sort(per + 1, per + m + 1, [&](int i, int j) { return c[i] < c[j]; });
    for (int j = 1; j <= (m); ++j) val[per[j]] = j;
    int dif = 0;
    for (int i = 0; i < (n); ++i) {
      int sc = 0;
      for (int j = 1; j <= (m); ++j)
        if (b[i][j]) sc += val[j];
      dif += abs(a[i] - sc);
    }
    if (dif > maxdif) {
      maxdif = dif;
      for (int j = 1; j <= (m); ++j) ans[j] = val[j];
    }
  }
  for (int j = 1; j <= (m); ++j) printf("%d%c", ans[j], " \n"[j == m]);
}

### 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;
int n, m;
int a[10], b[10][10005];
void Solve();
int main() {
  int tests = 1;
  scanf("%d", &tests);
  while (tests--) {
    Solve();
  }
  return 0;
}
void Solve() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < (n); ++i) scanf("%d", a + i);
  for (int i = 0; i < (n); ++i)
    for (int j = 1; j <= (m); ++j) scanf("%1d", &b[i][j]);
  static int ans[10005];
  int maxdif = -1;
  for (int s = 0; s < (1 << n); ++s) {
    static int c[10005];
    for (int j = 1; j <= (m); ++j) c[j] = 0;
    for (int i = 0; i < (n); ++i)
      for (int j = 1; j <= (m); ++j) c[j] += b[i][j] ? s >> i & 1 ? -1 : 1 : 0;
    static int per[10005], val[10005];
    for (int j = 1; j <= (m); ++j) per[j] = j;
    sort(per + 1, per + m + 1, [&](int i, int j) { return c[i] < c[j]; });
    for (int j = 1; j <= (m); ++j) val[per[j]] = j;
    int dif = 0;
    for (int i = 0; i < (n); ++i) {
      int sc = 0;
      for (int j = 1; j <= (m); ++j)
        if (b[i][j]) sc += val[j];
      dif += abs(a[i] - sc);
    }
    if (dif > maxdif) {
      maxdif = dif;
      for (int j = 1; j <= (m); ++j) ans[j] = val[j];
    }
  }
  for (int j = 1; j <= (m); ++j) printf("%d%c", ans[j], " \n"[j == m]);
}

```