#include <bits/stdc++.h>
using namespace std;
struct Node {
  long long num, id;
  inline bool operator<(const Node x) const {
    if (num != x.num) return num < x.num;
    return id < x.id;
  }
} a[100005];
long long T, n, m, Ans, sum, b[11], r[11], ans[100005];
char s[11][100005];
signed main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%d%d", &n, &m);
    for (long long i = 1; i <= n; i++) scanf("%d", &r[i]);
    for (long long i = 1; i <= n; i++) {
      scanf("%s", s[i] + 1);
    }
    Ans = -1e18;
    for (long long k = 0; k < (1 << n); k++) {
      long long now = k;
      sum = 0;
      for (long long i = n; i >= 1; i--) {
        b[i] = now % 2;
        if (b[i]) {
          sum -= r[i];
        } else {
          sum += r[i];
        }
        now /= 2;
      }
      for (long long i = 1; i <= m; i++) {
        a[i].id = i;
        a[i].num = 0;
        for (long long j = 1; j <= n; j++) {
          if (s[j][i] == '1') {
            if (b[j]) {
              a[i].num++;
            } else {
              a[i].num--;
            }
          }
        }
      }
      sort(a + 1, a + m + 1);
      for (long long i = 1; i <= m; i++) sum += a[i].num * i;
      if (sum > Ans) {
        Ans = sum;
        for (long long i = 1; i <= m; i++) ans[a[i].id] = i;
      }
    }
    for (long long i = 1; i <= m; i++) printf("%d ", ans[i]);
    puts("");
  }
  return 0;
}

### 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;
struct Node {
  long long num, id;
  inline bool operator<(const Node x) const {
    if (num != x.num) return num < x.num;
    return id < x.id;
  }
} a[100005];
long long T, n, m, Ans, sum, b[11], r[11], ans[100005];
char s[11][100005];
signed main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%d%d", &n, &m);
    for (long long i = 1; i <= n; i++) scanf("%d", &r[i]);
    for (long long i = 1; i <= n; i++) {
      scanf("%s", s[i] + 1);
    }
    Ans = -1e18;
    for (long long k = 0; k < (1 << n); k++) {
      long long now = k;
      sum = 0;
      for (long long i = n; i >= 1; i--) {
        b[i] = now % 2;
        if (b[i]) {
          sum -= r[i];
        } else {
          sum += r[i];
        }
        now /= 2;
      }
      for (long long i = 1; i <= m; i++) {
        a[i].id = i;
        a[i].num = 0;
        for (long long j = 1; j <= n; j++) {
          if (s[j][i] == '1') {
            if (b[j]) {
              a[i].num++;
            } else {
              a[i].num--;
            }
          }
        }
      }
      sort(a + 1, a + m + 1);
      for (long long i = 1; i <= m; i++) sum += a[i].num * i;
      if (sum > Ans) {
        Ans = sum;
        for (long long i = 1; i <= m; i++) ans[a[i].id] = i;
      }
    }
    for (long long i = 1; i <= m; i++) printf("%d ", ans[i]);
    puts("");
  }
  return 0;
}

```