#include <bits/stdc++.h>
using namespace std;
const long long N = 15, M = 1e4 + 1e2 + 7;
long long T;
long long n, m, x[N];
char s[N][M];
long long val[M], r[M], p[M], ap[M];
bool cmp(long long a, long long b) { return val[a] < val[b]; }
signed main() {
  scanf("%lld", &T);
  while (T--) {
    long long ans = -1;
    scanf("%lld%lld", &n, &m);
    for (long long i = 1; i <= n; i++) scanf("%lld", &x[i]);
    for (long long i = 1; i <= n; i++) scanf("%s", s[i] + 1);
    for (long long S = 0; S < (1 << n); S++) {
      for (long long i = 1; i <= m; i++) val[i] = 0, r[i] = i;
      for (long long i = 1; i <= n; i++) {
        long long v = (S & (1 << i - 1)) ? 1 : -1;
        for (long long j = 1; j <= m; j++)
          if (s[i][j] == '1') val[j] += v;
      }
      sort(r + 1, r + m + 1, cmp);
      for (long long i = 1; i <= m; i++) p[r[i]] = i;
      long long tot = 0;
      for (long long i = 1; i <= n; i++) {
        long long sv = 0;
        for (long long j = 1; j <= m; j++)
          if (s[i][j] == '1') sv += p[j];
        tot += abs(sv - x[i]);
      }
      if (tot > ans) {
        ans = tot;
        for (long long j = 1; j <= m; j++) ap[j] = p[j];
      }
    }
    for (long long i = 1; i <= m; i++) printf("%lld%c", ap[i], " \n"[i == m]);
  }
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 long long N = 15, M = 1e4 + 1e2 + 7;
long long T;
long long n, m, x[N];
char s[N][M];
long long val[M], r[M], p[M], ap[M];
bool cmp(long long a, long long b) { return val[a] < val[b]; }
signed main() {
  scanf("%lld", &T);
  while (T--) {
    long long ans = -1;
    scanf("%lld%lld", &n, &m);
    for (long long i = 1; i <= n; i++) scanf("%lld", &x[i]);
    for (long long i = 1; i <= n; i++) scanf("%s", s[i] + 1);
    for (long long S = 0; S < (1 << n); S++) {
      for (long long i = 1; i <= m; i++) val[i] = 0, r[i] = i;
      for (long long i = 1; i <= n; i++) {
        long long v = (S & (1 << i - 1)) ? 1 : -1;
        for (long long j = 1; j <= m; j++)
          if (s[i][j] == '1') val[j] += v;
      }
      sort(r + 1, r + m + 1, cmp);
      for (long long i = 1; i <= m; i++) p[r[i]] = i;
      long long tot = 0;
      for (long long i = 1; i <= n; i++) {
        long long sv = 0;
        for (long long j = 1; j <= m; j++)
          if (s[i][j] == '1') sv += p[j];
        tot += abs(sv - x[i]);
      }
      if (tot > ans) {
        ans = tot;
        for (long long j = 1; j <= m; j++) ap[j] = p[j];
      }
    }
    for (long long i = 1; i <= m; i++) printf("%lld%c", ap[i], " \n"[i == m]);
  }
}

```