#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
const int maxn = 1e4 + 10;
int n, m;
int a[maxn];
char s[12][maxn];
void solve() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < n; i++) {
    scanf("%s", s[i]);
  }
  ll ans = -1e18;
  vector<int> p(m);
  for (int b = 0; b < 1 << n; b++) {
    vector<int> f(n);
    ll now = 0;
    for (int i = 0; i < n; i++) {
      if (b >> i & 1)
        f[i] = 1;
      else
        f[i] = -1;
      if (f[i] == -1)
        now += a[i];
      else
        now -= a[i];
    }
    vector<pii> cost(m);
    for (int i = 0; i < m; i++) cost[i].second = i;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        if (s[i][j] != '0') cost[j].first += f[i];
      }
    }
    sort(cost.begin(), cost.end());
    for (int i = 0; i < m; i++) {
      now += 1ll * cost[i].first * (i + 1);
    }
    if (now >= ans) {
      ans = now;
      for (int i = 0; i < m; i++) {
        p[cost[i].second] = i + 1;
      }
    }
  }
  for (int i = 0; i < m; i++) printf("%d%c", p[i], " \n"[i == m - 1]);
}
int main() {
  int t;
  cin >> t;
  while (t--) solve();
}

### Prompt

In Cpp, your task is to solve 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;
using ll = long long;
using pii = pair<int, int>;
const int maxn = 1e4 + 10;
int n, m;
int a[maxn];
char s[12][maxn];
void solve() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < n; i++) {
    scanf("%s", s[i]);
  }
  ll ans = -1e18;
  vector<int> p(m);
  for (int b = 0; b < 1 << n; b++) {
    vector<int> f(n);
    ll now = 0;
    for (int i = 0; i < n; i++) {
      if (b >> i & 1)
        f[i] = 1;
      else
        f[i] = -1;
      if (f[i] == -1)
        now += a[i];
      else
        now -= a[i];
    }
    vector<pii> cost(m);
    for (int i = 0; i < m; i++) cost[i].second = i;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        if (s[i][j] != '0') cost[j].first += f[i];
      }
    }
    sort(cost.begin(), cost.end());
    for (int i = 0; i < m; i++) {
      now += 1ll * cost[i].first * (i + 1);
    }
    if (now >= ans) {
      ans = now;
      for (int i = 0; i < m; i++) {
        p[cost[i].second] = i + 1;
      }
    }
  }
  for (int i = 0; i < m; i++) printf("%d%c", p[i], " \n"[i == m - 1]);
}
int main() {
  int t;
  cin >> t;
  while (t--) solve();
}

```