#include <bits/stdc++.h>
using namespace std;
const int N = 12;
const int M = 10010;
int T, n, m, x[N], w[M], ans[M];
char s[N][M];
pair<int, int> a[M];
int main() {
  scanf("%d", &T);
  for (int t = 1; t <= T; t++) {
    scanf("%d %d", &n, &m);
    for (int i = 0; i < n; i++) {
      scanf("%d", &x[i]);
    }
    for (int i = 0; i < n; i++) {
      scanf("%s", s[i]);
    }
    for (int i = 0; i < m; i++) {
      w[i] = 0;
      for (int j = 0; j < n; j++) {
        w[i] |= ((s[j][i] == '1') ? (1 << j) : 0);
      }
    }
    long long res = -1;
    for (int i = 0; i < (1 << n); i++) {
      int neg = 0;
      int pos = 0;
      long long sum = 0;
      for (int j = 0; j < n; j++) {
        if ((i & (1 << j)) == 0) {
          neg |= (1 << j);
          sum += x[j];
        } else {
          pos |= (1 << j);
          sum -= x[j];
        }
      }
      for (int j = 0; j < m; j++) {
        int mneg = (neg & w[j]);
        int mpos = (pos & w[j]);
        int cnt = __builtin_popcount(mpos) - __builtin_popcount(mneg);
        a[j] = {cnt, j + 1};
      }
      sort(a, a + m);
      for (int j = 0; j < m; j++) {
        sum += a[j].first * (j + 1);
      }
      if (sum > res) {
        res = sum;
        for (int j = 0; j < m; j++) {
          ans[a[j].second - 1] = j + 1;
        }
      }
    }
    for (int i = 0; i < m; i++) {
      printf("%d ", ans[i]);
    }
    puts("");
  }
  return 0;
}

### 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 = 12;
const int M = 10010;
int T, n, m, x[N], w[M], ans[M];
char s[N][M];
pair<int, int> a[M];
int main() {
  scanf("%d", &T);
  for (int t = 1; t <= T; t++) {
    scanf("%d %d", &n, &m);
    for (int i = 0; i < n; i++) {
      scanf("%d", &x[i]);
    }
    for (int i = 0; i < n; i++) {
      scanf("%s", s[i]);
    }
    for (int i = 0; i < m; i++) {
      w[i] = 0;
      for (int j = 0; j < n; j++) {
        w[i] |= ((s[j][i] == '1') ? (1 << j) : 0);
      }
    }
    long long res = -1;
    for (int i = 0; i < (1 << n); i++) {
      int neg = 0;
      int pos = 0;
      long long sum = 0;
      for (int j = 0; j < n; j++) {
        if ((i & (1 << j)) == 0) {
          neg |= (1 << j);
          sum += x[j];
        } else {
          pos |= (1 << j);
          sum -= x[j];
        }
      }
      for (int j = 0; j < m; j++) {
        int mneg = (neg & w[j]);
        int mpos = (pos & w[j]);
        int cnt = __builtin_popcount(mpos) - __builtin_popcount(mneg);
        a[j] = {cnt, j + 1};
      }
      sort(a, a + m);
      for (int j = 0; j < m; j++) {
        sum += a[j].first * (j + 1);
      }
      if (sum > res) {
        res = sum;
        for (int j = 0; j < m; j++) {
          ans[a[j].second - 1] = j + 1;
        }
      }
    }
    for (int i = 0; i < m; i++) {
      printf("%d ", ans[i]);
    }
    puts("");
  }
  return 0;
}

```