#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    static int xx[10], ans[10000];
    static char s[10][10000 + 1];
    int n, m, ansv;
    scanf("%d%d", &n, &m);
    for (int i = 0; i < n; i++) scanf("%d", &xx[i]);
    for (int i = 0; i < n; i++) scanf("%s", s[i]);
    ansv = -1;
    memset(ans, 0, m * sizeof *ans);
    for (int b = 0; b < 1 << n; b++) {
      static int kk[10000];
      int sum = 0, cnt;
      vector<int> ii(m);
      for (int i = 0; i < n; i++)
        if (b >> i & 1)
          sum += xx[i];
        else
          sum -= xx[i];
      for (int h = 0; h < m; h++) {
        int k = 0;
        for (int i = 0; i < n; i++)
          if (s[i][h] == '1') {
            if (b >> i & 1)
              k--;
            else
              k++;
          }
        kk[h] = k, ii[h] = h;
      }
      sort(ii.begin(), ii.end(), [&](int i, int j) { return kk[i] < kk[j]; });
      cnt = 1;
      for (int h : ii) sum += kk[h] * cnt++;
      if (sum > ansv) {
        ansv = sum;
        cnt = 1;
        for (int h : ii) {
          ans[h] = cnt++;
        }
      }
    }
    for (int h = 0; h < m; h++) printf("%d ", ans[h]);
    printf("\n");
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    static int xx[10], ans[10000];
    static char s[10][10000 + 1];
    int n, m, ansv;
    scanf("%d%d", &n, &m);
    for (int i = 0; i < n; i++) scanf("%d", &xx[i]);
    for (int i = 0; i < n; i++) scanf("%s", s[i]);
    ansv = -1;
    memset(ans, 0, m * sizeof *ans);
    for (int b = 0; b < 1 << n; b++) {
      static int kk[10000];
      int sum = 0, cnt;
      vector<int> ii(m);
      for (int i = 0; i < n; i++)
        if (b >> i & 1)
          sum += xx[i];
        else
          sum -= xx[i];
      for (int h = 0; h < m; h++) {
        int k = 0;
        for (int i = 0; i < n; i++)
          if (s[i][h] == '1') {
            if (b >> i & 1)
              k--;
            else
              k++;
          }
        kk[h] = k, ii[h] = h;
      }
      sort(ii.begin(), ii.end(), [&](int i, int j) { return kk[i] < kk[j]; });
      cnt = 1;
      for (int h : ii) sum += kk[h] * cnt++;
      if (sum > ansv) {
        ansv = sum;
        cnt = 1;
        for (int h : ii) {
          ans[h] = cnt++;
        }
      }
    }
    for (int h = 0; h < m; h++) printf("%d ", ans[h]);
    printf("\n");
  }
  return 0;
}

```