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;
clock_t start;
mt19937_64 rng(chrono::system_clock::now().time_since_epoch().count());
template <typename T>
void _F(const char *name, T arg1) {
  cerr << name << " = " << arg1 << '\n';
}
template <typename T, typename... Args>
void _F(const char *names, T arg1, Args... args) {
  const char *name = strchr(names, ',');
  cerr.write(names, name - names) << " = " << arg1 << '\n';
  _F(name + 2, args...);
}
template <typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &q) {
  in >> q.first >> q.second;
  return in;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &out, pair<T1, T2> &q) {
  out << q.first << " " << q.second;
  return out;
}
template <typename T1, typename T2>
pair<T1, T2> operator+(pair<T1, T2> p1, pair<T1, T2> p2) {
  return {p1.first + p2.first, p1.second + p2.second};
}
template <typename T1, typename T2>
pair<T1, T2> operator-(pair<T1, T2> p1, pair<T1, T2> p2) {
  return {p1.first - p2.first, p1.second - p2.second};
}
template <typename T1, typename T2>
bool operator<(pair<T1, T2> p1, pair<T1, T2> p2) {
  return p1 < p2;
}
template <typename T>
void Unique(vector<T> &v) {
  sort((v).begin(), (v).end()),
      v.resize(distance(v.begin(), unique((v).begin(), (v).end())));
}
void solve(long long &test) {
  long long n;
  cin >> n;
  string a, b;
  cin >> a >> b;
  std::vector<long long> ans;
  long long sum = 0, cnt = 0;
  auto invert = [&](char c, long long cnt) {
    if (cnt % 2 == 0) return c;
    return (c == '1' ? '0' : '1');
  };
  for (long long i = n - 1; i >= 0; i--) {
    long long j = sum + (cnt % 2 ? -i : i);
    char cur = a[j];
    cur = invert(cur, cnt);
    if (b[i] == cur) continue;
    if (cur == invert(a[sum], cnt)) {
      ans.push_back(i);
    } else {
      a[sum] = invert(a[sum], 1);
      ans.push_back(0);
      ans.push_back(i);
    }
    sum += (cnt % 2 == 0 ? i : -i);
    cnt++;
  }
  cout << (long long)((ans).size()) << '\n';
  for (auto u : ans) cout << u + 1 << " ";
  cout << '\n';
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  start = clock();
  long long t = 1;
  cin >> t;
  cout << fixed << setprecision(2);
  for (long long i = 1; i <= t; ++i) {
    solve(i);
  }
  cerr << (double(clock() - start) / double(CLOCKS_PER_SEC));
  return 0;
}