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;
template <class TH>
void _dbg(const char *sdbg, TH h) {
  cerr << sdbg << '=' << h << endl;
}
template <class TH, class... TA>
void _dbg(const char *sdbg, TH h, TA... a) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << '=' << h << ',';
  _dbg(sdbg + 1, a...);
}
template <class T>
ostream &operator<<(ostream &os, vector<T> V) {
  os << "[";
  for (auto vv : V) os << vv << ",";
  return os << "]";
}
template <class L, class R>
ostream &operator<<(ostream &os, pair<L, R> P) {
  return os << "(" << P.first << "," << P.second << ")";
}
template <class T>
using min_heap = priority_queue<T, vector<T>, greater<T>>;
template <class T>
using max_heap = priority_queue<T>;
const int mod = 998244353;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
struct PairHash {
  template <typename T1, typename T2>
  std::size_t operator()(const pair<T1, T2> &p) const {
    return hash<T1>()(p.first) ^ hash<T2>()(p.second);
  }
};
inline int flap(int c) { return (1 - ((int)c - '0')) + (int)'0'; }
void solve(int ncase) {
  int n;
  cin >> n;
  string a, b;
  cin >> a >> b;
  vector<int> ret;
  for (int i = a.size() - 1; i >= 0; i--) {
    if (a[i] == b[i]) continue;
    if (i == 0) {
      a[0] = flap(a[0]);
      ret.push_back(1);
      continue;
    }
    if (b[i] == a[0]) {
      ret.push_back(1);
      a[0] = flap(a[0]);
    }
    ret.push_back(i + 1);
    for (int j = i; j >= 0; j--) a[j] = flap(a[j]);
    for (int k = 0, j = i; k <= j; k++, j--) swap(a[k], a[j]);
  }
  cout << ret.size();
  for (auto x : ret) cout << " " << x;
  cout << endl;
}
void solve_all_cases() {
  int T = 1;
  cin >> T;
  int ncase = 0;
  while (T--) {
    solve(++ncase);
  }
}
int main() {
  cout << std::fixed;
  cout << setprecision(9);
  solve_all_cases();
}