#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using db = long double;
const int maxn = 2e5 + 5;
const ll mod = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int CASOS;
  cin >> CASOS;
  for (int caso = 1; caso <= CASOS; caso++) {
    int n, m;
    cin >> n >> m;
    vector<ll> x(n);
    for (auto &i : x) cin >> i;
    vector<pair<int, int>> c(m);
    vector<string> s(n);
    for (auto &i : s) cin >> i;
    vector<int> p(m), aux(m);
    ll ans = LLONG_MIN;
    for (int mask = 0; mask < (1 << n); mask++) {
      for (int j = 0; j < m; j++) {
        c[j] = {0, j};
        for (int i = 0; i < n; i++) {
          if (mask & (1 << i))
            c[j].first -= (s[i][j] - '0');
          else
            c[j].first += (s[i][j] - '0');
        }
      }
      sort((c).begin(), (c).end());
      ll cur = 0;
      for (int i = 0; i < m; i++) {
        aux[c[i].second] = i + 1;
        cur += (ll)(i + 1) * (ll)c[i].first;
      }
      for (int i = 0; i < n; i++) {
        if (mask & (1 << i))
          cur += x[i];
        else
          cur -= x[i];
      }
      if (ans < cur) {
        ans = cur;
        p = aux;
      }
    }
    for (auto i : p) cout << i << ' ';
    cout << '\n';
  }
  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;
using db = long double;
const int maxn = 2e5 + 5;
const ll mod = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int CASOS;
  cin >> CASOS;
  for (int caso = 1; caso <= CASOS; caso++) {
    int n, m;
    cin >> n >> m;
    vector<ll> x(n);
    for (auto &i : x) cin >> i;
    vector<pair<int, int>> c(m);
    vector<string> s(n);
    for (auto &i : s) cin >> i;
    vector<int> p(m), aux(m);
    ll ans = LLONG_MIN;
    for (int mask = 0; mask < (1 << n); mask++) {
      for (int j = 0; j < m; j++) {
        c[j] = {0, j};
        for (int i = 0; i < n; i++) {
          if (mask & (1 << i))
            c[j].first -= (s[i][j] - '0');
          else
            c[j].first += (s[i][j] - '0');
        }
      }
      sort((c).begin(), (c).end());
      ll cur = 0;
      for (int i = 0; i < m; i++) {
        aux[c[i].second] = i + 1;
        cur += (ll)(i + 1) * (ll)c[i].first;
      }
      for (int i = 0; i < n; i++) {
        if (mask & (1 << i))
          cur += x[i];
        else
          cur -= x[i];
      }
      if (ans < cur) {
        ans = cur;
        p = aux;
      }
    }
    for (auto i : p) cout << i << ' ';
    cout << '\n';
  }
  return 0;
}

```