#include <bits/stdc++.h>
using namespace std;
int t, n, m, a[15], ans, b[10010];
char s[15][10010];
struct P {
  int val, id;
} p[10010];
inline bool cmp(P x, P y) { return x.val < y.val; }
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d%d", &n, &m);
    ans = 0;
    for (int i = 0; i < n; i++) scanf("%d", &a[i]);
    for (int i = 0; i < n; i++) scanf("%s", s[i] + 1);
    for (int i = 0; i < 1 << n; i++) {
      int sum = 0;
      for (int j = 1; j <= m; j++) {
        p[j].val = 0;
        p[j].id = j;
      }
      for (int j = 0; j < n; j++) {
        if (i & (1 << j)) {
          sum += a[j];
          for (int k = 1; k <= m; k++)
            if (s[j][k] == '1') p[k].val--;
        } else {
          sum -= a[j];
          for (int k = 1; k <= m; k++)
            if (s[j][k] == '1') p[k].val++;
        }
      }
      sort(p + 1, p + m + 1, cmp);
      for (int j = 1; j <= m; j++) sum += j * p[j].val;
      if (sum >= ans) {
        ans = sum;
        for (int j = 1; j <= m; j++) b[p[j].id] = j;
      }
    }
    for (int i = 1; i <= m; i++) printf("%d ", b[i]);
    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>
using namespace std;
int t, n, m, a[15], ans, b[10010];
char s[15][10010];
struct P {
  int val, id;
} p[10010];
inline bool cmp(P x, P y) { return x.val < y.val; }
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d%d", &n, &m);
    ans = 0;
    for (int i = 0; i < n; i++) scanf("%d", &a[i]);
    for (int i = 0; i < n; i++) scanf("%s", s[i] + 1);
    for (int i = 0; i < 1 << n; i++) {
      int sum = 0;
      for (int j = 1; j <= m; j++) {
        p[j].val = 0;
        p[j].id = j;
      }
      for (int j = 0; j < n; j++) {
        if (i & (1 << j)) {
          sum += a[j];
          for (int k = 1; k <= m; k++)
            if (s[j][k] == '1') p[k].val--;
        } else {
          sum -= a[j];
          for (int k = 1; k <= m; k++)
            if (s[j][k] == '1') p[k].val++;
        }
      }
      sort(p + 1, p + m + 1, cmp);
      for (int j = 1; j <= m; j++) sum += j * p[j].val;
      if (sum >= ans) {
        ans = sum;
        for (int j = 1; j <= m; j++) b[p[j].id] = j;
      }
    }
    for (int i = 1; i <= m; i++) printf("%d ", b[i]);
    printf("\n");
  }
  return 0;
}

```