#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int ans = 0, f = 1;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    ans = (ans << 1) + (ans << 3) + c - '0';
    c = getchar();
  }
  return ans * f;
}
inline void write(int x) {
  if (x < 0) putchar('-'), x = -x;
  if (x / 10) write(x / 10);
  putchar((char)(x % 10) + '0');
}
template <typename T>
inline T Abs(T a) {
  return a > 0 ? a : -a;
};
template <typename T, typename TT>
inline T Min(T a, TT b) {
  return a < b ? a : b;
}
template <typename T, typename TT>
inline T Max(T a, TT b) {
  return a < b ? b : a;
}
const int N = 1e4 + 5;
int t, n, m, a[N], c[N], ans;
char s[15][N];
pair<int, int> b[N];
signed main() {
  t = read();
  while (t--) {
    n = read();
    m = read();
    for (int i = 1; i <= n; ++i) a[i] = read();
    for (int i = 1; i <= n; ++i) scanf("%s", s[i] + 1);
    ans = -1e9;
    for (int i = 0; i < (1 << n); ++i) {
      for (int j = 1; j <= m; ++j) {
        b[j] = make_pair(0, j);
        for (int k = 1; k <= n; ++k)
          if (s[k][j] == '1') {
            if ((i >> (k - 1)) & 1)
              b[j].first++;
            else
              b[j].first--;
          }
      }
      sort(b + 1, b + 1 + m);
      int tmp = 0;
      for (int j = 1; j <= n; ++j)
        if ((i >> (j - 1)) & 1)
          tmp -= a[j];
        else
          tmp += a[j];
      for (int j = 1; j <= m; ++j) tmp += j * b[j].first;
      if (tmp > ans) {
        ans = tmp;
        for (int j = 1; j <= m; ++j) c[b[j].second] = j;
      }
    }
    for (int i = 1; i <= m; ++i) write(c[i]), putchar(' ');
    puts("");
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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>
using namespace std;
inline int read() {
  int ans = 0, f = 1;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    ans = (ans << 1) + (ans << 3) + c - '0';
    c = getchar();
  }
  return ans * f;
}
inline void write(int x) {
  if (x < 0) putchar('-'), x = -x;
  if (x / 10) write(x / 10);
  putchar((char)(x % 10) + '0');
}
template <typename T>
inline T Abs(T a) {
  return a > 0 ? a : -a;
};
template <typename T, typename TT>
inline T Min(T a, TT b) {
  return a < b ? a : b;
}
template <typename T, typename TT>
inline T Max(T a, TT b) {
  return a < b ? b : a;
}
const int N = 1e4 + 5;
int t, n, m, a[N], c[N], ans;
char s[15][N];
pair<int, int> b[N];
signed main() {
  t = read();
  while (t--) {
    n = read();
    m = read();
    for (int i = 1; i <= n; ++i) a[i] = read();
    for (int i = 1; i <= n; ++i) scanf("%s", s[i] + 1);
    ans = -1e9;
    for (int i = 0; i < (1 << n); ++i) {
      for (int j = 1; j <= m; ++j) {
        b[j] = make_pair(0, j);
        for (int k = 1; k <= n; ++k)
          if (s[k][j] == '1') {
            if ((i >> (k - 1)) & 1)
              b[j].first++;
            else
              b[j].first--;
          }
      }
      sort(b + 1, b + 1 + m);
      int tmp = 0;
      for (int j = 1; j <= n; ++j)
        if ((i >> (j - 1)) & 1)
          tmp -= a[j];
        else
          tmp += a[j];
      for (int j = 1; j <= m; ++j) tmp += j * b[j].first;
      if (tmp > ans) {
        ans = tmp;
        for (int j = 1; j <= m; ++j) c[b[j].second] = j;
      }
    }
    for (int i = 1; i <= m; ++i) write(c[i]), putchar(' ');
    puts("");
  }
  return 0;
}

```