#include <bits/stdc++.h>
using namespace std;
const int N = 11, M = 1e4 + 5;
int n, m, p[M];
long long res;
struct Student {
  int p, q[M];
} a[N];
struct Problem {
  int cnt, id;
} b[M];
bool cmp(Problem a, Problem b) { return a.cnt < b.cnt; }
int solve() {
  cin >> n >> m, res = -1;
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i].p);
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) scanf("%1d", &a[i].q[j]);
  }
  for (int i = 0; i < (1 << n); ++i) {
    for (int j = 1; j <= m; ++j) {
      b[j].id = j, b[j].cnt = 0;
      for (int k = 0; k < n; ++k)
        b[j].cnt += a[k + 1].q[j] * (((i >> k) & 1) ? 1 : -1);
    }
    sort(b + 1, b + 1 + m, cmp);
    long long sum = 0;
    for (int j = 0; j < n; ++j) sum += a[j + 1].p * (((i >> j) & 1) ? -1 : 1);
    for (int j = 1; j <= m; ++j) sum += b[j].cnt * j;
    if (sum > res) {
      res = sum;
      for (int j = 1; j <= m; ++j) p[b[j].id] = j;
    }
  }
  for (int i = 1; i <= m; ++i) printf("%d ", p[i]);
  return printf("\n"), 0;
}
int main() {
  int T;
  cin >> T;
  while (T--) solve();
}

### Prompt

Generate 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;
const int N = 11, M = 1e4 + 5;
int n, m, p[M];
long long res;
struct Student {
  int p, q[M];
} a[N];
struct Problem {
  int cnt, id;
} b[M];
bool cmp(Problem a, Problem b) { return a.cnt < b.cnt; }
int solve() {
  cin >> n >> m, res = -1;
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i].p);
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) scanf("%1d", &a[i].q[j]);
  }
  for (int i = 0; i < (1 << n); ++i) {
    for (int j = 1; j <= m; ++j) {
      b[j].id = j, b[j].cnt = 0;
      for (int k = 0; k < n; ++k)
        b[j].cnt += a[k + 1].q[j] * (((i >> k) & 1) ? 1 : -1);
    }
    sort(b + 1, b + 1 + m, cmp);
    long long sum = 0;
    for (int j = 0; j < n; ++j) sum += a[j + 1].p * (((i >> j) & 1) ? -1 : 1);
    for (int j = 1; j <= m; ++j) sum += b[j].cnt * j;
    if (sum > res) {
      res = sum;
      for (int j = 1; j <= m; ++j) p[b[j].id] = j;
    }
  }
  for (int i = 1; i <= m; ++i) printf("%d ", p[i]);
  return printf("\n"), 0;
}
int main() {
  int T;
  cin >> T;
  while (T--) solve();
}

```