#include <bits/stdc++.h>
#pragma GCC optimize("O3,no-stack-protector,unroll-loops")
#pragma GCC target("avx2,fma")
using namespace std;
const long long inf = 1e18, N = 1e4 + 10, maxn = 11, mod = 1e9 + 7,
                pr = 1000696969;
int x[maxn], st[N], dp[N];
vector<pair<long long, int> > vc;
string s[maxn];
inline void solve() {
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < n; i++) cin >> x[i];
  for (int i = 0; i < n; i++) cin >> s[i];
  long long ans = 0, ans_t = -inf;
  for (int mask = 0; mask < (1 << n); mask++) {
    vc.clear();
    ans = 0;
    for (int j = 0; j < m; j++) dp[j] = 0;
    for (int j = 0; j < n; j++) {
      if (mask & (1 << j)) {
        ans += x[j];
        for (int k = 0; k < m; k++) {
          if (s[j][k] == '1') {
            dp[k]--;
          }
        }
      } else {
        ans -= x[j];
        for (int k = 0; k < m; k++) {
          if (s[j][k] == '1') {
            dp[k]++;
          }
        }
      }
    }
    for (int i = 0; i < m; i++) {
      vc.push_back({dp[i], i});
    }
    sort((vc).begin(), (vc).end());
    for (int i = int((vc).size()) - 1; i >= 0; i--) {
      ans += 1ll * vc[i].first * (i + 1);
    }
    if (ans > ans_t) {
      for (int i = int((vc).size()) - 1; i >= 0; i--) {
        st[vc[i].second] = i + 1;
      }
      ans_t = ans;
    }
  }
  for (int i = 0; i < m; i++) cout << st[i] << " ";
  cout << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}

### 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("O3,no-stack-protector,unroll-loops")
#pragma GCC target("avx2,fma")
using namespace std;
const long long inf = 1e18, N = 1e4 + 10, maxn = 11, mod = 1e9 + 7,
                pr = 1000696969;
int x[maxn], st[N], dp[N];
vector<pair<long long, int> > vc;
string s[maxn];
inline void solve() {
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < n; i++) cin >> x[i];
  for (int i = 0; i < n; i++) cin >> s[i];
  long long ans = 0, ans_t = -inf;
  for (int mask = 0; mask < (1 << n); mask++) {
    vc.clear();
    ans = 0;
    for (int j = 0; j < m; j++) dp[j] = 0;
    for (int j = 0; j < n; j++) {
      if (mask & (1 << j)) {
        ans += x[j];
        for (int k = 0; k < m; k++) {
          if (s[j][k] == '1') {
            dp[k]--;
          }
        }
      } else {
        ans -= x[j];
        for (int k = 0; k < m; k++) {
          if (s[j][k] == '1') {
            dp[k]++;
          }
        }
      }
    }
    for (int i = 0; i < m; i++) {
      vc.push_back({dp[i], i});
    }
    sort((vc).begin(), (vc).end());
    for (int i = int((vc).size()) - 1; i >= 0; i--) {
      ans += 1ll * vc[i].first * (i + 1);
    }
    if (ans > ans_t) {
      for (int i = int((vc).size()) - 1; i >= 0; i--) {
        st[vc[i].second] = i + 1;
      }
      ans_t = ans;
    }
  }
  for (int i = 0; i < m; i++) cout << st[i] << " ";
  cout << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}

```