#include <bits/stdc++.h>
using namespace std;
const int N = 12;
const int M = 10005;
const int lim = 200000;
const int inf = 1e9;
int T;
int n, m;
int x[N];
char s[N][M];
int cnt[M];
int b[M];
int ans[M];
bool cmp(int i, int j) { return cnt[i] < cnt[j]; }
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%d%d", &n, &m);
    for (int i = 0; i <= n - 1; ++i) scanf("%d", &x[i]);
    for (int i = 0; i <= n - 1; ++i) scanf("%s", s[i]);
    int ret = -inf;
    for (int mask = 0; mask <= (1 << n) - 1; ++mask) {
      for (int j = 0; j <= m - 1; ++j) cnt[j] = 0;
      for (int i = 0; i <= n - 1; ++i) {
        for (int j = 0; j <= m - 1; ++j) {
          if (s[i][j] == '1') {
            if (mask & (1 << i))
              ++cnt[j];
            else
              --cnt[j];
          }
        }
      }
      for (int j = 0; j <= m - 1; ++j) b[j] = j;
      sort(b, b + m, cmp);
      int tmp = 0;
      for (int i = 0; i <= n - 1; ++i) {
        int r = 0;
        for (int j = 0; j <= m - 1; ++j) {
          if (s[i][b[j]] == '1') {
            r += j + 1;
          }
        }
        if (mask & (1 << i))
          tmp += r - x[i];
        else
          tmp -= r - x[i];
      }
      if (tmp > ret) {
        ret = tmp;
        for (int j = 0; j <= m - 1; ++j) ans[b[j]] = j + 1;
      }
    }
    for (int j = 0; j <= m - 1; ++j) printf("%d ", ans[j]);
    printf("\n");
  }
  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 = 10005;
const int lim = 200000;
const int inf = 1e9;
int T;
int n, m;
int x[N];
char s[N][M];
int cnt[M];
int b[M];
int ans[M];
bool cmp(int i, int j) { return cnt[i] < cnt[j]; }
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%d%d", &n, &m);
    for (int i = 0; i <= n - 1; ++i) scanf("%d", &x[i]);
    for (int i = 0; i <= n - 1; ++i) scanf("%s", s[i]);
    int ret = -inf;
    for (int mask = 0; mask <= (1 << n) - 1; ++mask) {
      for (int j = 0; j <= m - 1; ++j) cnt[j] = 0;
      for (int i = 0; i <= n - 1; ++i) {
        for (int j = 0; j <= m - 1; ++j) {
          if (s[i][j] == '1') {
            if (mask & (1 << i))
              ++cnt[j];
            else
              --cnt[j];
          }
        }
      }
      for (int j = 0; j <= m - 1; ++j) b[j] = j;
      sort(b, b + m, cmp);
      int tmp = 0;
      for (int i = 0; i <= n - 1; ++i) {
        int r = 0;
        for (int j = 0; j <= m - 1; ++j) {
          if (s[i][b[j]] == '1') {
            r += j + 1;
          }
        }
        if (mask & (1 << i))
          tmp += r - x[i];
        else
          tmp -= r - x[i];
      }
      if (tmp > ret) {
        ret = tmp;
        for (int j = 0; j <= m - 1; ++j) ans[b[j]] = j + 1;
      }
    }
    for (int j = 0; j <= m - 1; ++j) printf("%d ", ans[j]);
    printf("\n");
  }
  return 0;
}

```