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> v;
string s1;
void reverse(long long i) {
  long long k = i;
  v.push_back(i + 1);
  for (int j = 0; j < (i % 2 + i / 2); j++) {
    if (s1[j] == '0')
      s1[j] = '1';
    else
      s1[j] = '0';
    if (s1[k] == '0')
      s1[k] = '1';
    else
      s1[k] = '0';
    swap(s1[j], s1[k]);
    k--;
  }
  if (i % 2 == 0 && s1[i / 2] == '1')
    s1[i / 2] = '0';
  else if (i % 2 == 0 && s1[i / 2] == '0')
    s1[i / 2] = '1';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    string s2;
    cin >> s1 >> s2;
    for (int i = (n - 1); i >= 0; i--) {
      if (s1[i] == s2[i]) continue;
      if (s1[0] != s2[i])
        reverse(i);
      else {
        v.push_back(1);
        if (s1[0] == '0')
          s1[0] = '1';
        else
          s1[0] = '0';
        reverse(i);
      }
    }
    cout << v.size() << " ";
    for (auto &e : v) cout << e << " ";
    cout << "\n";
    v.clear();
  }
  return 0;
}