#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC target("popcnt")
using namespace std;
mt19937 bruh(chrono::steady_clock::now().time_since_epoch().count());
const long long N = 1e6 + 100;
const long long mod = 998244353;
const long long big = 1e17;
long long t, n, m;
string a[N];
pair<long long, long long> c[N];
int x[N], d[N], e[N];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> t;
  while (t--) {
    cin >> n >> m;
    for (int i = 0; i < n; i++) cin >> x[i];
    for (int i = 0; i < n; i++) cin >> a[i];
    long long ans = -1e10;
    for (int j = 0; j < (1 << n); j++) {
      for (int i = 0; i < m; i++) c[i] = make_pair(0, i), d[i] = 0;
      long long sum = 0;
      for (int i = 0; i < n; i++) {
        if ((1 << i) & j) {
          sum += x[i];
          for (int k = 0; k < m; k++) {
            if (a[i][k] == '1') c[k].first--;
          }
        } else {
          sum -= x[i];
          for (int k = 0; k < m; k++)
            if (a[i][k] == '1') c[k].first++;
        }
      }
      sort(c, c + m);
      for (int i = 0; i < m; i++)
        d[c[i].second] = i + 1, sum = sum + c[i].first * (i + 1);
      if ((sum > ans)) {
        ans = sum;
        for (int j = 0; j < m; j++) e[j] = d[j];
      }
    }
    for (int i = 0; i < m; i++) cout << e[i] << " ";
    cout << '\n';
  }
}

### Prompt

Develop a solution in Cpp to the problem described below:

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")
#pragma GCC target("popcnt")
using namespace std;
mt19937 bruh(chrono::steady_clock::now().time_since_epoch().count());
const long long N = 1e6 + 100;
const long long mod = 998244353;
const long long big = 1e17;
long long t, n, m;
string a[N];
pair<long long, long long> c[N];
int x[N], d[N], e[N];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> t;
  while (t--) {
    cin >> n >> m;
    for (int i = 0; i < n; i++) cin >> x[i];
    for (int i = 0; i < n; i++) cin >> a[i];
    long long ans = -1e10;
    for (int j = 0; j < (1 << n); j++) {
      for (int i = 0; i < m; i++) c[i] = make_pair(0, i), d[i] = 0;
      long long sum = 0;
      for (int i = 0; i < n; i++) {
        if ((1 << i) & j) {
          sum += x[i];
          for (int k = 0; k < m; k++) {
            if (a[i][k] == '1') c[k].first--;
          }
        } else {
          sum -= x[i];
          for (int k = 0; k < m; k++)
            if (a[i][k] == '1') c[k].first++;
        }
      }
      sort(c, c + m);
      for (int i = 0; i < m; i++)
        d[c[i].second] = i + 1, sum = sum + c[i].first * (i + 1);
      if ((sum > ans)) {
        ans = sum;
        for (int j = 0; j < m; j++) e[j] = d[j];
      }
    }
    for (int i = 0; i < m; i++) cout << e[i] << " ";
    cout << '\n';
  }
}

```