#include <bits/stdc++.h>
using namespace std;
int T, n, m;
int x[15];
char s[15][10005];
int v[10005], cnt[25], tmp[25];
int P[10005];
int ans;
void dfs(int p, int sum) {
  if (p == n + 1) {
    int cur = m, res = sum;
    for (int i = 20; i >= 0; i--) {
      res += (i - 10) * (cur * 2 - cnt[i] + 1) * cnt[i] / 2;
      cur -= cnt[i];
    }
    tmp[0] = cnt[0];
    for (int i = 1; i <= 20; i++) tmp[i] = tmp[i - 1] + cnt[i];
    if (ans <= res) {
      ans = res;
      for (int i = 1; i <= m; i++) {
        P[i] = tmp[v[i] + 10]--;
      }
    }
    return;
  }
  for (int i = 1; i <= m; i++) {
    if (s[p][i] == '1') {
      cnt[v[i] + 10]--;
      v[i]++;
      cnt[v[i] + 10]++;
    }
  }
  dfs(p + 1, sum - x[p]);
  for (int i = 1; i <= m; i++) {
    if (s[p][i] == '1') {
      cnt[v[i] + 10]--;
      v[i] -= 2;
      cnt[v[i] + 10]++;
    }
  }
  dfs(p + 1, sum + x[p]);
  for (int i = 1; i <= m; i++) {
    if (s[p][i] == '1') {
      cnt[v[i] + 10]--;
      v[i]++;
      cnt[v[i] + 10]++;
    }
  }
}
void solve() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &x[i]);
  for (int i = 1; i <= n; i++) {
    scanf("%s", &s[i][1]);
  }
  for (int i = 1; i <= m; i++) v[i] = 0;
  for (int i = 0; i <= 20; i++) cnt[i] = 0;
  cnt[10] = m;
  ans = 0;
  dfs(1, 0);
  for (int i = 1; i <= m; i++) printf("%d ", P[i]);
  printf("\n");
}
int main() {
  scanf("%d", &T);
  while (T--) solve();
  return 0;
}

### Prompt

Your task is to create 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;
int x[15];
char s[15][10005];
int v[10005], cnt[25], tmp[25];
int P[10005];
int ans;
void dfs(int p, int sum) {
  if (p == n + 1) {
    int cur = m, res = sum;
    for (int i = 20; i >= 0; i--) {
      res += (i - 10) * (cur * 2 - cnt[i] + 1) * cnt[i] / 2;
      cur -= cnt[i];
    }
    tmp[0] = cnt[0];
    for (int i = 1; i <= 20; i++) tmp[i] = tmp[i - 1] + cnt[i];
    if (ans <= res) {
      ans = res;
      for (int i = 1; i <= m; i++) {
        P[i] = tmp[v[i] + 10]--;
      }
    }
    return;
  }
  for (int i = 1; i <= m; i++) {
    if (s[p][i] == '1') {
      cnt[v[i] + 10]--;
      v[i]++;
      cnt[v[i] + 10]++;
    }
  }
  dfs(p + 1, sum - x[p]);
  for (int i = 1; i <= m; i++) {
    if (s[p][i] == '1') {
      cnt[v[i] + 10]--;
      v[i] -= 2;
      cnt[v[i] + 10]++;
    }
  }
  dfs(p + 1, sum + x[p]);
  for (int i = 1; i <= m; i++) {
    if (s[p][i] == '1') {
      cnt[v[i] + 10]--;
      v[i]++;
      cnt[v[i] + 10]++;
    }
  }
}
void solve() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &x[i]);
  for (int i = 1; i <= n; i++) {
    scanf("%s", &s[i][1]);
  }
  for (int i = 1; i <= m; i++) v[i] = 0;
  for (int i = 0; i <= 20; i++) cnt[i] = 0;
  cnt[10] = m;
  ans = 0;
  dfs(1, 0);
  for (int i = 1; i <= m; i++) printf("%d ", P[i]);
  printf("\n");
}
int main() {
  scanf("%d", &T);
  while (T--) solve();
  return 0;
}

```