#include <bits/stdc++.h>
using namespace std;
using LL = long long;
const int MAXN = 11;
const int MAXM = 1e4 + 5;
int n, m, ex[MAXN];
int cnt[30];
char w[MAXN][MAXM];
void solve() {
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; i++) scanf("%d", &ex[i]);
  for (int i = 0; i < n; i++) scanf("%s", w[i]);
  LL ans = 0;
  int pos;
  for (int i = 0; i < (1 << n); i++) {
    for (int j = 0; j <= 20; j++) cnt[j] = 0;
    for (int j = 0; j < m; j++) {
      int x[2] = {0, 0};
      for (int k = 0; k < n; k++)
        if (w[k][j] == '1') x[((1 << k) & i) != 0]++;
      cnt[x[1] - x[0] + 10]++;
    }
    int mm = m;
    LL sum = 0;
    for (int j = 0; j < n; j++)
      if (((1 << j) & i) == 0)
        sum += ex[j];
      else
        sum -= ex[j];
    for (int j = 20; j >= 0; j--) {
      sum += 1ll * (j - 10) * (2 * mm - cnt[j] + 1) * cnt[j] / 2;
      mm -= cnt[j];
    }
    if (sum >= ans) ans = sum, pos = i;
  }
  {
    for (int j = 0; j <= 20; j++) cnt[j] = 0;
    for (int j = 0; j < m; j++) {
      int x[2] = {0, 0};
      for (int k = 0; k < n; k++)
        if (w[k][j] == '1') x[((1 << k) & pos) != 0]++;
      cnt[x[1] - x[0] + 10]++;
    }
    for (int j = 1; j <= 20; j++) cnt[j] += cnt[j - 1];
    for (int j = 0; j < m; j++) {
      int x[2] = {0, 0};
      for (int k = 0; k < n; k++)
        if (w[k][j] == '1') x[((1 << k) & pos) != 0]++;
      printf("%d ", cnt[x[1] - x[0] + 10]);
      cnt[x[1] - x[0] + 10]--;
    }
    printf("\n");
  }
}
int main() {
  int T;
  scanf("%d", &T);
  for (int i = 1; i <= T; i++) {
    solve();
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
using LL = long long;
const int MAXN = 11;
const int MAXM = 1e4 + 5;
int n, m, ex[MAXN];
int cnt[30];
char w[MAXN][MAXM];
void solve() {
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; i++) scanf("%d", &ex[i]);
  for (int i = 0; i < n; i++) scanf("%s", w[i]);
  LL ans = 0;
  int pos;
  for (int i = 0; i < (1 << n); i++) {
    for (int j = 0; j <= 20; j++) cnt[j] = 0;
    for (int j = 0; j < m; j++) {
      int x[2] = {0, 0};
      for (int k = 0; k < n; k++)
        if (w[k][j] == '1') x[((1 << k) & i) != 0]++;
      cnt[x[1] - x[0] + 10]++;
    }
    int mm = m;
    LL sum = 0;
    for (int j = 0; j < n; j++)
      if (((1 << j) & i) == 0)
        sum += ex[j];
      else
        sum -= ex[j];
    for (int j = 20; j >= 0; j--) {
      sum += 1ll * (j - 10) * (2 * mm - cnt[j] + 1) * cnt[j] / 2;
      mm -= cnt[j];
    }
    if (sum >= ans) ans = sum, pos = i;
  }
  {
    for (int j = 0; j <= 20; j++) cnt[j] = 0;
    for (int j = 0; j < m; j++) {
      int x[2] = {0, 0};
      for (int k = 0; k < n; k++)
        if (w[k][j] == '1') x[((1 << k) & pos) != 0]++;
      cnt[x[1] - x[0] + 10]++;
    }
    for (int j = 1; j <= 20; j++) cnt[j] += cnt[j - 1];
    for (int j = 0; j < m; j++) {
      int x[2] = {0, 0};
      for (int k = 0; k < n; k++)
        if (w[k][j] == '1') x[((1 << k) & pos) != 0]++;
      printf("%d ", cnt[x[1] - x[0] + 10]);
      cnt[x[1] - x[0] + 10]--;
    }
    printf("\n");
  }
}
int main() {
  int T;
  scanf("%d", &T);
  for (int i = 1; i <= T; i++) {
    solve();
  }
  return 0;
}

```