#include <bits/stdc++.h>
int ww[10000];
void sort(int *jj, int m) {
  static int kk[(10 + 1) * 2], jj_[10000];
  int j, w;
  memset(kk, 0, sizeof kk);
  for (j = 0; j < m; j++) {
    w = 10 + ww[j];
    kk[w + 1]++;
  }
  for (w = 1; w < (10 + 1) * 2; w++) kk[w] += kk[w - 1];
  for (j = 0; j < m; j++) {
    w = 10 + ww[j];
    jj_[kk[w]++] = jj[j];
  }
  memcpy(jj, jj_, m * sizeof *jj_);
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    static int xx[10], jj[10000], jj_[10000], pp[10000];
    static char cc[10][10000 + 1];
    int n, m, i, j, b, sum_, sum;
    scanf("%d%d", &n, &m);
    for (i = 0; i < n; i++) scanf("%d", &xx[i]);
    for (i = 0; i < n; i++) scanf("%s", cc[i]);
    sum_ = -1;
    for (b = 0; b < 1 << n; b++) {
      memset(ww, 0, m * sizeof *ww);
      sum = 0;
      for (i = 0; i < n; i++)
        if ((b & 1 << i) == 0) {
          sum += xx[i];
          for (j = 0; j < m; j++)
            if (cc[i][j] == '1') ww[j]--;
        } else {
          sum -= xx[i];
          for (j = 0; j < m; j++)
            if (cc[i][j] == '1') ww[j]++;
        }
      for (j = 0; j < m; j++) jj[j] = j;
      sort(jj, m);
      for (j = 0; j < m; j++) sum += ww[jj[j]] * (j + 1);
      if (sum_ < sum) {
        sum_ = sum;
        memcpy(jj_, jj, m * sizeof *jj);
      }
    }
    for (j = 0; j < m; j++) pp[jj_[j]] = j + 1;
    for (j = 0; j < m; j++) printf("%d ", pp[j]);
    printf("\n");
  }
  return 0;
}

### Prompt

Please formulate 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>
int ww[10000];
void sort(int *jj, int m) {
  static int kk[(10 + 1) * 2], jj_[10000];
  int j, w;
  memset(kk, 0, sizeof kk);
  for (j = 0; j < m; j++) {
    w = 10 + ww[j];
    kk[w + 1]++;
  }
  for (w = 1; w < (10 + 1) * 2; w++) kk[w] += kk[w - 1];
  for (j = 0; j < m; j++) {
    w = 10 + ww[j];
    jj_[kk[w]++] = jj[j];
  }
  memcpy(jj, jj_, m * sizeof *jj_);
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    static int xx[10], jj[10000], jj_[10000], pp[10000];
    static char cc[10][10000 + 1];
    int n, m, i, j, b, sum_, sum;
    scanf("%d%d", &n, &m);
    for (i = 0; i < n; i++) scanf("%d", &xx[i]);
    for (i = 0; i < n; i++) scanf("%s", cc[i]);
    sum_ = -1;
    for (b = 0; b < 1 << n; b++) {
      memset(ww, 0, m * sizeof *ww);
      sum = 0;
      for (i = 0; i < n; i++)
        if ((b & 1 << i) == 0) {
          sum += xx[i];
          for (j = 0; j < m; j++)
            if (cc[i][j] == '1') ww[j]--;
        } else {
          sum -= xx[i];
          for (j = 0; j < m; j++)
            if (cc[i][j] == '1') ww[j]++;
        }
      for (j = 0; j < m; j++) jj[j] = j;
      sort(jj, m);
      for (j = 0; j < m; j++) sum += ww[jj[j]] * (j + 1);
      if (sum_ < sum) {
        sum_ = sum;
        memcpy(jj_, jj, m * sizeof *jj);
      }
    }
    for (j = 0; j < m; j++) pp[jj_[j]] = j + 1;
    for (j = 0; j < m; j++) printf("%d ", pp[j]);
    printf("\n");
  }
  return 0;
}

```