Problem: This is the easy version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.

There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.

For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.

Your task is to transform the string a into b in at most 3n operations. It can be proved that it is always possible.

Input

The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.

The first line of each test case contains a single integer n (1≤ n≤ 1000) — the length of the binary strings.

The next two lines contain two binary strings a and b of length n.

It is guaranteed that the sum of n across all test cases does not exceed 1000.

Output

For each test case, output an integer k (0≤ k≤ 3n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.

Example

Input


5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1


Output


3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1

Note

In the first test case, we have 01→ 11→ 00→ 10.

In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.

In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
vector<long long int> fact(501), in(501);
long long int v1[110][110], v2[110][110];
long long int powy(long long int a, long long int b) {
  if (b == 0) return 1;
  long long int c = powy(a, b / 2);
  if (b & 1)
    return (c * ((c * a) % 998244353) % 998244353);
  else
    return ((c * c) % 998244353);
}
long long int c(long long int n, long long int r) {
  return fact[n] * in[r] % 998244353 * in[n - r] % 998244353;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    string s1, s2;
    cin >> s1 >> s2;
    vector<long long int> ans;
    long long int x = 0, i = n - 1;
    while (x < (4 * n)) {
      if (s1[0] == s2[i]) {
        ans.push_back(1);
        s1[0] == '0' ? s1[0] = '1' : s1[0] = '0';
      }
      ans.push_back(i + 1);
      string temp;
      for (long long int j = 0; j < i + 1; j++) temp = temp + s1[j];
      for (long long int j = 0; j < i + 1; j++)
        temp[j] == '0' ? temp[j] = '1' : temp[j] = '0';
      reverse(temp.begin(), temp.end());
      for (long long int j = 0; j < i + 1; j++) s1[j] = temp[j];
      x++;
      i--;
      if (i < 0) break;
    }
    long long int d = ans.size();
    cout << d << " ";
    for (long long int i = 0; i < d; i++) cout << ans[i] << " ";
    cout << "\n";
  }
  return 0;
}