#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int n, m;
long long res = -1;
string t[20];
long long w[20];
int ans[200000];
void get(long long x) {
  int s = 0;
  int k[n], f[n], p = 0;
  for (int i = 0; i < n; i++) k[i] = 1;
  while (x > 0) {
    if (x % 2 == 1) {
      k[p] = -1;
      s += w[p];
    }
    p++;
    x /= 2;
  }
  for (int i = 0; i < n; i++)
    if (k[i] == 1) s -= w[i];
  vector<pair<int, int> > a(m);
  for (int i = 0; i < m; i++) {
    a[i].first = 0;
    a[i].second = i;
  }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++)
      if (t[i][j] == '1') a[j].first += k[i];
  sort(a.begin(), a.end());
  for (int i = 0; i < m; i++) s += a[i].first * (i + 1);
  if (s > res) {
    res = s;
    for (int i = 0; i < m; i++) ans[a[i].second] = i + 1;
    return;
  }
}
void solve() {
  cin >> n >> m;
  for (int i = 0; i < n; i++) cin >> w[i];
  for (int i = 0; i < n; i++) cin >> t[i];
  res = -1;
  for (int x = 0; x < (1 << n); x++) get(x);
  for (int i = 0; i < m; i++) cout << ans[i] << ' ';
  cout << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  int q;
  cin >> q;
  while (q--) solve();
}

### Prompt

Your challenge is to write 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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int n, m;
long long res = -1;
string t[20];
long long w[20];
int ans[200000];
void get(long long x) {
  int s = 0;
  int k[n], f[n], p = 0;
  for (int i = 0; i < n; i++) k[i] = 1;
  while (x > 0) {
    if (x % 2 == 1) {
      k[p] = -1;
      s += w[p];
    }
    p++;
    x /= 2;
  }
  for (int i = 0; i < n; i++)
    if (k[i] == 1) s -= w[i];
  vector<pair<int, int> > a(m);
  for (int i = 0; i < m; i++) {
    a[i].first = 0;
    a[i].second = i;
  }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++)
      if (t[i][j] == '1') a[j].first += k[i];
  sort(a.begin(), a.end());
  for (int i = 0; i < m; i++) s += a[i].first * (i + 1);
  if (s > res) {
    res = s;
    for (int i = 0; i < m; i++) ans[a[i].second] = i + 1;
    return;
  }
}
void solve() {
  cin >> n >> m;
  for (int i = 0; i < n; i++) cin >> w[i];
  for (int i = 0; i < n; i++) cin >> t[i];
  res = -1;
  for (int x = 0; x < (1 << n); x++) get(x);
  for (int i = 0; i < m; i++) cout << ans[i] << ' ';
  cout << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  int q;
  cin >> q;
  while (q--) solve();
}

```