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.
#include <bits/stdc++.h>
using namespace std;
inline long long _gcd(long long a, long long b) {
  while (b) b %= a ^= b ^= a ^= b;
  return a;
}
inline pair<int, int> operator+(pair<int, int> a, pair<int, int> b) {
  return {a.first + b.first, a.second + b.second};
}
inline ostream& operator<<(ostream& o, pair<int, int> p) {
  o << p.first << " " << p.second;
  return o;
}
template <typename>
struct is_std_vector : std::false_type {};
template <typename T, typename A>
struct is_std_vector<std::vector<T, A>> : std::true_type {};
template <typename T>
inline ostream& operator<<(ostream& o, vector<T>& p) {
  for (int i = 0; i < p.size(); ++i) {
    o << setw(3) << p[i];
    if (is_std_vector<T>::value) o << "\n";
  }
  return o;
}
string s1, s2;
int n;
void swapStr(int len) {
  string s3 = s1;
  for (int i = 0; i < len; ++i) {
    if (s3[len - i - 1] == '1')
      s1[i] = '0';
    else
      s1[i] = '1';
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    cin >> n;
    cin >> s1;
    cin >> s2;
    vector<int> swaps;
    swaps.reserve(2 * n);
    for (int i = n - 1; i >= 0; --i) {
      if (s1[0] == s2[i]) {
        swaps.push_back(1);
        swapStr(1);
      }
      swaps.push_back(i + 1);
      swapStr(i + 1);
    }
    cout << swaps.size();
    for (auto a : swaps) {
      cout << " " << a;
    }
    cout << "\n";
  }
  return 0;
}
