#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e4 + 9;
const long long MOD = 1e9 + 7;
const int mod = 998244353;
inline long long qpow(long long b, long long e, long long m) {
  long long a = 1;
  for (; e; e >>= 1, b = b * b % m)
    if (e & 1) a = a * b % m;
  return a;
}
long long exgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1, y = 0;
    return a;
  }
  long long d = exgcd(b, a % b, x, y);
  long long z = x;
  x = y, y = z - y * (a / b);
  return d;
}
int x[19];
char s[19][maxn];
int ans[maxn];
pair<int, int> val[maxn];
bool cmp(pair<int, int> A, pair<int, int> B) { return A.second < B.second; }
int main() {
  int _;
  scanf("%d", &_);
  while (_--) {
    int n, m;
    scanf("%d%d", &n, &m);
    for (int i = 0; i < n; i++) scanf("%d", &x[i]);
    for (int i = 0; i < n; i++) {
      scanf("%s", s[i]);
    }
    int v = -1;
    for (int i = 0; i < (1 << n); i++) {
      int res = 0;
      for (int j = 0; j < n; j++) {
        if (i >> j & 1)
          res += x[j];
        else
          res -= x[j];
      }
      for (int k = 0; k < m; k++) {
        val[k].first = 0;
        val[k].second = k;
      }
      for (int j = 0; j < n; j++) {
        for (int k = 0; k < m; k++) {
          if (s[j][k] == '1') {
            if (i >> j & 1)
              val[k].first--;
            else
              val[k].first++;
          }
        }
      }
      sort(val, val + m);
      for (int j = 0; j < m; j++) {
        res += (j + 1) * val[j].first;
      }
      if (res > v) {
        v = res;
        for (int j = 0; j < m; j++) {
          ans[val[j].second] = j + 1;
        }
      }
    }
    for (int i = 0; i < m; i++) {
      printf("%d ", ans[i]);
    }
    printf("\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>
using namespace std;
const int maxn = 1e4 + 9;
const long long MOD = 1e9 + 7;
const int mod = 998244353;
inline long long qpow(long long b, long long e, long long m) {
  long long a = 1;
  for (; e; e >>= 1, b = b * b % m)
    if (e & 1) a = a * b % m;
  return a;
}
long long exgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1, y = 0;
    return a;
  }
  long long d = exgcd(b, a % b, x, y);
  long long z = x;
  x = y, y = z - y * (a / b);
  return d;
}
int x[19];
char s[19][maxn];
int ans[maxn];
pair<int, int> val[maxn];
bool cmp(pair<int, int> A, pair<int, int> B) { return A.second < B.second; }
int main() {
  int _;
  scanf("%d", &_);
  while (_--) {
    int n, m;
    scanf("%d%d", &n, &m);
    for (int i = 0; i < n; i++) scanf("%d", &x[i]);
    for (int i = 0; i < n; i++) {
      scanf("%s", s[i]);
    }
    int v = -1;
    for (int i = 0; i < (1 << n); i++) {
      int res = 0;
      for (int j = 0; j < n; j++) {
        if (i >> j & 1)
          res += x[j];
        else
          res -= x[j];
      }
      for (int k = 0; k < m; k++) {
        val[k].first = 0;
        val[k].second = k;
      }
      for (int j = 0; j < n; j++) {
        for (int k = 0; k < m; k++) {
          if (s[j][k] == '1') {
            if (i >> j & 1)
              val[k].first--;
            else
              val[k].first++;
          }
        }
      }
      sort(val, val + m);
      for (int j = 0; j < m; j++) {
        res += (j + 1) * val[j].first;
      }
      if (res > v) {
        v = res;
        for (int j = 0; j < m; j++) {
          ans[val[j].second] = j + 1;
        }
      }
    }
    for (int i = 0; i < m; i++) {
      printf("%d ", ans[i]);
    }
    printf("\n");
  }
}

```