#include <bits/stdc++.h>
using namespace std;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long n, m;
    cin >> n >> m;
    vector<long long> x(n);
    for (long long i = 0; i < n; ++i) {
      cin >> x[i];
    }
    vector<string> s(n);
    for (long long i = 0; i < n; ++i) {
      cin >> s[i];
    }
    vector<vector<long long>> gr(1 << n);
    for (long long j = 0; j < m; ++j) {
      long long sum = 0;
      for (long long i = 0; i < n; ++i) {
        sum = (sum << 1) + s[i][j] - '0';
      }
      gr[sum].push_back(j);
    }
    vector<vector<long long>> srt;
    for (long long i = 0; i < (1 << n); ++i) {
      if ((long long)gr[i].size()) {
        srt.push_back(gr[i]);
      }
    }
    vector<long long> col(m);
    for (long long i = 0; i < (long long)srt.size(); ++i) {
      for (auto& j : srt[i]) {
        col[j] = i;
      }
    }
    vector<vector<long long>> uses(n);
    for (long long i = 0; i < n; ++i) {
      for (long long j = 0; j < m; ++j) {
        if (s[i][j] == '1') {
          uses[i].push_back(col[j]);
        }
      }
      sort(uses[i].begin(), uses[i].end());
      uses[i].erase(unique(uses[i].begin(), uses[i].end()), uses[i].end());
    }
    long long ans = 0;
    vector<long long> vans;
    for (long long i = 0; i < (1 << n); ++i) {
      long long sum = 0;
      vector<long long> gcnt((long long)srt.size());
      for (long long j = 0; j < n; ++j) {
        if (i & (1 << j)) {
          sum += x[j];
          for (auto& x : uses[j]) {
            --gcnt[x];
          }
        } else {
          sum -= x[j];
          for (auto& x : uses[j]) {
            ++gcnt[x];
          }
        }
      }
      vector<vector<long long>> gsrt;
      for (long long h = 0; h < (long long)srt.size(); ++h) {
        gsrt.push_back({gcnt[h], (long long)srt[h].size(), h});
      }
      sort(gsrt.begin(), gsrt.end());
      long long l = 1;
      for (long long h = 0; h < (long long)srt.size(); ++h) {
        long long r = l + (long long)gsrt[h][1] - 1;
        sum += (r + l) * (r - l + 1) / 2 * gsrt[h][0];
        l = r + 1;
      }
      if (sum >= ans) {
        ans = sum;
        vans.clear();
        for (long long h = 0; h < (long long)srt.size(); ++h) {
          vans.push_back(gsrt[h][2]);
        }
      }
    }
    vector<long long> out(m);
    long long l = 1;
    for (auto& i : vans) {
      for (auto& j : srt[i]) {
        out[j] = l++;
      }
    }
    for (long long i = 0; i < m; ++i) {
      cout << out[i] << ' ';
    }
    cout << '\n';
  }
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long n, m;
    cin >> n >> m;
    vector<long long> x(n);
    for (long long i = 0; i < n; ++i) {
      cin >> x[i];
    }
    vector<string> s(n);
    for (long long i = 0; i < n; ++i) {
      cin >> s[i];
    }
    vector<vector<long long>> gr(1 << n);
    for (long long j = 0; j < m; ++j) {
      long long sum = 0;
      for (long long i = 0; i < n; ++i) {
        sum = (sum << 1) + s[i][j] - '0';
      }
      gr[sum].push_back(j);
    }
    vector<vector<long long>> srt;
    for (long long i = 0; i < (1 << n); ++i) {
      if ((long long)gr[i].size()) {
        srt.push_back(gr[i]);
      }
    }
    vector<long long> col(m);
    for (long long i = 0; i < (long long)srt.size(); ++i) {
      for (auto& j : srt[i]) {
        col[j] = i;
      }
    }
    vector<vector<long long>> uses(n);
    for (long long i = 0; i < n; ++i) {
      for (long long j = 0; j < m; ++j) {
        if (s[i][j] == '1') {
          uses[i].push_back(col[j]);
        }
      }
      sort(uses[i].begin(), uses[i].end());
      uses[i].erase(unique(uses[i].begin(), uses[i].end()), uses[i].end());
    }
    long long ans = 0;
    vector<long long> vans;
    for (long long i = 0; i < (1 << n); ++i) {
      long long sum = 0;
      vector<long long> gcnt((long long)srt.size());
      for (long long j = 0; j < n; ++j) {
        if (i & (1 << j)) {
          sum += x[j];
          for (auto& x : uses[j]) {
            --gcnt[x];
          }
        } else {
          sum -= x[j];
          for (auto& x : uses[j]) {
            ++gcnt[x];
          }
        }
      }
      vector<vector<long long>> gsrt;
      for (long long h = 0; h < (long long)srt.size(); ++h) {
        gsrt.push_back({gcnt[h], (long long)srt[h].size(), h});
      }
      sort(gsrt.begin(), gsrt.end());
      long long l = 1;
      for (long long h = 0; h < (long long)srt.size(); ++h) {
        long long r = l + (long long)gsrt[h][1] - 1;
        sum += (r + l) * (r - l + 1) / 2 * gsrt[h][0];
        l = r + 1;
      }
      if (sum >= ans) {
        ans = sum;
        vans.clear();
        for (long long h = 0; h < (long long)srt.size(); ++h) {
          vans.push_back(gsrt[h][2]);
        }
      }
    }
    vector<long long> out(m);
    long long l = 1;
    for (auto& i : vans) {
      for (auto& j : srt[i]) {
        out[j] = l++;
      }
    }
    for (long long i = 0; i < m; ++i) {
      cout << out[i] << ' ';
    }
    cout << '\n';
  }
  return 0;
}

```