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;
using ll = long long;
constexpr ll INF = (ll)1e18;
constexpr ll MOD = (ll)1e9 + 7;
constexpr double EPS = 1e-9;
constexpr ll dx[4] = {1, 0, -1, 0};
constexpr ll dy[4] = {0, 1, 0, -1};
constexpr ll dx8[8] = {1, 0, -1, 0, 1, 1, -1, -1};
constexpr ll dy8[8] = {0, 1, 0, -1, 1, -1, 1, -1};
void p() { std::cout << '\n'; }
template <typename Head, typename... Tail>
void p(Head head, Tail... tail) {
  std::cout << head << (sizeof...(tail) ? " " : "");
  p(tail...);
}
template <typename T>
inline void pv(std::vector<T>& v) {
  for (ll i = 0, N = v.size(); i < N; i++)
    std::cout << v[i] << " \n"[i == N - 1];
}
template <typename T>
inline bool chmax(T& a, T b) {
  return a < b && (a = b, true);
}
template <typename T>
inline bool chmin(T& a, T b) {
  return a > b && (a = b, true);
}
template <typename T>
inline void uniq(std::vector<T>& v) {
  v.erase(std::unique(v.begin(), v.end()), v.end());
}
void solve() {
  ll N;
  cin >> N;
  string A, B;
  cin >> A >> B;
  vector<ll> ans;
  for (ll i = 0, i_length = (N); i < i_length; ++i) {
    if (A[N - 1 - i] == B[N - 1 - i]) continue;
    if (i == N - 1) {
      if (A[0] != B[0]) ans.push_back(1);
      break;
    }
    if (A[0] == B[N - 1 - i]) {
      ans.push_back(1);
      A[0] = B[N - 1 - i];
    }
    string t = A.substr(0, N - i);
    string s = A.substr(N - i, i);
    reverse((t).begin(), (t).end());
    for (ll i = 0, i_length = (t.size()); i < i_length; ++i)
      if (t[i] == '0')
        t[i] = '1';
      else
        t[i] = '0';
    A = t + s;
    ans.push_back(N - i);
  }
  cout << ans.size() << ' ';
  pv(ans);
}
signed main() {
  std::cin.tie(nullptr);
  std::ios::sync_with_stdio(false);
  ll Q;
  cin >> Q;
  while (Q--) solve();
  return 0;
}
