#include <bits/stdc++.h>
using namespace std;
template <typename T>
bool chkMax(T &x, T y) {
  return (y > x) ? x = y, 1 : 0;
}
template <typename T>
bool chkMin(T &x, T y) {
  return (y < x) ? x = y, 1 : 0;
}
template <typename T>
void inline read(T &x) {
  int f = 1;
  x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s <= '9' && s >= '0') x = x * 10 + (s ^ 48), s = getchar();
  x *= f;
}
const int N = 10, M = 10000;
int n, m, ans[M];
pair<int, int> c[M];
long long x[N], val;
int s[N][M];
int main() {
  int T;
  read(T);
  while (T--) {
    val = -9e18;
    read(n), read(m);
    for (int i = 0; i < n; i++) read(x[i]);
    for (int i = 0; i < n; i++)
      for (int j = 0; j < m; j++) scanf("%1d", &s[i][j]);
    for (int i = 0; i < (1 << n); i++) {
      long long ret = 0;
      for (int j = 0; j < m; j++) c[j].first = 0, c[j].second = j;
      for (int j = 0; j < n; j++) {
        int t = (i >> j & 1) ? 1 : -1;
        for (int k = 0; k < m; k++) {
          if (s[j][k]) c[k].first += t;
        }
        if (t == 1)
          ret -= x[j];
        else
          ret += x[j];
      }
      sort(c, c + m);
      for (int j = 0; j < m; j++) ret += c[j].first * (j + 1ll);
      if (chkMax(val, ret)) {
        for (int j = 0; j < m; j++) ans[c[j].second] = j;
      }
    }
    for (int i = 0; i < m; i++) printf("%d ", ans[i] + 1);
    puts("");
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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;
template <typename T>
bool chkMax(T &x, T y) {
  return (y > x) ? x = y, 1 : 0;
}
template <typename T>
bool chkMin(T &x, T y) {
  return (y < x) ? x = y, 1 : 0;
}
template <typename T>
void inline read(T &x) {
  int f = 1;
  x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s <= '9' && s >= '0') x = x * 10 + (s ^ 48), s = getchar();
  x *= f;
}
const int N = 10, M = 10000;
int n, m, ans[M];
pair<int, int> c[M];
long long x[N], val;
int s[N][M];
int main() {
  int T;
  read(T);
  while (T--) {
    val = -9e18;
    read(n), read(m);
    for (int i = 0; i < n; i++) read(x[i]);
    for (int i = 0; i < n; i++)
      for (int j = 0; j < m; j++) scanf("%1d", &s[i][j]);
    for (int i = 0; i < (1 << n); i++) {
      long long ret = 0;
      for (int j = 0; j < m; j++) c[j].first = 0, c[j].second = j;
      for (int j = 0; j < n; j++) {
        int t = (i >> j & 1) ? 1 : -1;
        for (int k = 0; k < m; k++) {
          if (s[j][k]) c[k].first += t;
        }
        if (t == 1)
          ret -= x[j];
        else
          ret += x[j];
      }
      sort(c, c + m);
      for (int j = 0; j < m; j++) ret += c[j].first * (j + 1ll);
      if (chkMax(val, ret)) {
        for (int j = 0; j < m; j++) ans[c[j].second] = j;
      }
    }
    for (int i = 0; i < m; i++) printf("%d ", ans[i] + 1);
    puts("");
  }
  return 0;
}

```