#include <bits/stdc++.h>
using namespace std;
const int MAXN = 10 + 5;
const int MAXM = 1e4 + 5;
const long long linf = 0x3f3f3f3f3f3f3f3f;
int n, m;
int a[MAXN];
char s[MAXN][MAXM];
int p[MAXM];
inline long long get_res(int mask) {
  long long has = 0;
  static int coef[MAXM];
  memset(coef, 0, (m + 1) << 2);
  for (int i = 1; i <= n; ++i) {
    int t = (((mask) >> (i - 1)) & 1) ? 1 : -1;
    has += -t * a[i];
    for (int j = 1; j <= m; ++j)
      if (s[i][j] == '1') coef[j] += t;
  }
  static int id[MAXM];
  for (int i = 1; i <= m; ++i) id[i] = i;
  sort(id + 1, id + m + 1, [](int x, int y) { return coef[x] < coef[y]; });
  for (int i = 1; i <= m; ++i) p[id[i]] = i;
  for (int i = 1; i <= m; ++i) has += coef[i] * p[i];
  return has;
}
void solve(void) {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= n; ++i) scanf("%s", s[i] + 1);
  int all = (1 << n) - 1;
  long long ans = -linf;
  int ansmask = -1;
  for (int mask = 0; mask <= all; ++mask) {
    long long cur = get_res(mask);
    if (cur > ans) ans = cur, ansmask = mask;
  }
  get_res(ansmask);
  for (int i = 1; i <= m; ++i) printf("%d ", p[i]);
  printf("\n");
}
int main(void) {
  int T;
  scanf("%d", &T);
  while (T--) solve();
  return 0;
}

### Prompt

In cpp, your task is to solve 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 MAXN = 10 + 5;
const int MAXM = 1e4 + 5;
const long long linf = 0x3f3f3f3f3f3f3f3f;
int n, m;
int a[MAXN];
char s[MAXN][MAXM];
int p[MAXM];
inline long long get_res(int mask) {
  long long has = 0;
  static int coef[MAXM];
  memset(coef, 0, (m + 1) << 2);
  for (int i = 1; i <= n; ++i) {
    int t = (((mask) >> (i - 1)) & 1) ? 1 : -1;
    has += -t * a[i];
    for (int j = 1; j <= m; ++j)
      if (s[i][j] == '1') coef[j] += t;
  }
  static int id[MAXM];
  for (int i = 1; i <= m; ++i) id[i] = i;
  sort(id + 1, id + m + 1, [](int x, int y) { return coef[x] < coef[y]; });
  for (int i = 1; i <= m; ++i) p[id[i]] = i;
  for (int i = 1; i <= m; ++i) has += coef[i] * p[i];
  return has;
}
void solve(void) {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= n; ++i) scanf("%s", s[i] + 1);
  int all = (1 << n) - 1;
  long long ans = -linf;
  int ansmask = -1;
  for (int mask = 0; mask <= all; ++mask) {
    long long cur = get_res(mask);
    if (cur > ans) ans = cur, ansmask = mask;
  }
  get_res(ansmask);
  for (int i = 1; i <= m; ++i) printf("%d ", p[i]);
  printf("\n");
}
int main(void) {
  int T;
  scanf("%d", &T);
  while (T--) solve();
  return 0;
}

```