#include <bits/stdc++.h>
#pragma GCC optimize("O4,unroll-loops,no-stack-protector")
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
using namespace std;
int n, m;
int arr[13][10005];
int e[13];
vector<int> ans(10005), Ans(10005);
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    cin >> n >> m;
    for (int i = 0; i < n; i++) cin >> e[i];
    for (int i = 0; i < n; i++) {
      for (int j = 1; j <= m; j++) {
        char c;
        cin >> c;
        arr[i][j] = (c == '1');
      }
    }
    int best = -1;
    for (int i = 0; i <= m; i++) {
      ans[i] = 0;
      Ans[i] = 0;
    }
    for (int bit = 0; bit < (1 << n); bit++) {
      vector<pair<long long, long long> > cnt1(m + 2), cnt2(m + 2);
      cnt1[0] = make_pair(-1, -1);
      cnt2[0] = make_pair(-1, -1);
      for (int i = 1; i <= m; i++) {
        cnt1[i].second = i;
        cnt2[i].second = i;
      }
      for (int i = 0; i < n; i++) {
        for (int j = 1; j <= m; j++)
          if (arr[i][j]) {
            if ((1 << i) & bit)
              cnt1[j].first++;
            else
              cnt1[j].first--;
          }
      }
      sort(cnt1.begin() + 1, cnt1.begin() + 1 + m);
      for (int j = 1; j <= m; j++) {
        ans[cnt1[j].second] = j;
      }
      int res = 0;
      for (int i = 0; i < n; i++) {
        int now = 0;
        for (int j = 1; j <= m; j++)
          if (arr[i][j]) now += ans[j];
        res += abs(now - e[i]);
      }
      if (res > best) {
        best = res;
        Ans = ans;
      }
    }
    for (int i = 1; i <= m; i++) cout << Ans[i] << " ";
    cout << "\n";
  }
}

### Prompt

Your task is to create 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("O4,unroll-loops,no-stack-protector")
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
using namespace std;
int n, m;
int arr[13][10005];
int e[13];
vector<int> ans(10005), Ans(10005);
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    cin >> n >> m;
    for (int i = 0; i < n; i++) cin >> e[i];
    for (int i = 0; i < n; i++) {
      for (int j = 1; j <= m; j++) {
        char c;
        cin >> c;
        arr[i][j] = (c == '1');
      }
    }
    int best = -1;
    for (int i = 0; i <= m; i++) {
      ans[i] = 0;
      Ans[i] = 0;
    }
    for (int bit = 0; bit < (1 << n); bit++) {
      vector<pair<long long, long long> > cnt1(m + 2), cnt2(m + 2);
      cnt1[0] = make_pair(-1, -1);
      cnt2[0] = make_pair(-1, -1);
      for (int i = 1; i <= m; i++) {
        cnt1[i].second = i;
        cnt2[i].second = i;
      }
      for (int i = 0; i < n; i++) {
        for (int j = 1; j <= m; j++)
          if (arr[i][j]) {
            if ((1 << i) & bit)
              cnt1[j].first++;
            else
              cnt1[j].first--;
          }
      }
      sort(cnt1.begin() + 1, cnt1.begin() + 1 + m);
      for (int j = 1; j <= m; j++) {
        ans[cnt1[j].second] = j;
      }
      int res = 0;
      for (int i = 0; i < n; i++) {
        int now = 0;
        for (int j = 1; j <= m; j++)
          if (arr[i][j]) now += ans[j];
        res += abs(now - e[i]);
      }
      if (res > best) {
        best = res;
        Ans = ans;
      }
    }
    for (int i = 1; i <= m; i++) cout << Ans[i] << " ";
    cout << "\n";
  }
}

```