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;
const long long md = 1e9 + 7;
const long long MX = 2e5 + 5;
const long long INF = 1e18;
const long double PI = 4 * atan((long double)1);
long long power_md(long long a, long long n) {
  long long res = 1;
  while (n) {
    if (n % 2) res = (res % md * a % md) % md;
    a = (a % md * a % md) % md;
    n /= 2;
  }
  res %= md;
  return res;
}
long long power(long long a, long long n) {
  long long res = 1;
  while (n) {
    if (n % 2) res *= a;
    a = a * a;
    n /= 2;
  }
  return res;
}
long long abst(long long a) { return ((a < 0) ? (-1 * a) : (a)); }
class cmp_set {
 public:
  bool operator()(long long a, long long b) { return a > b; }
};
vector<string> vec_splitter(string s) {
  s += ',';
  vector<string> res;
  while (!s.empty()) {
    res.push_back(s.substr(0, s.find(',')));
    s = s.substr(s.find(',') + 1);
  }
  return res;
}
void debug_out(vector<string> __attribute__((unused)) args,
               __attribute__((unused)) long long idx,
               __attribute__((unused)) long long LINE_NUM) {
  cerr << endl;
}
template <typename Head, typename... Tail>
void debug_out(vector<string> args, long long idx, long long LINE_NUM, Head H,
               Tail... T) {
  if (idx > 0)
    cerr << ", ";
  else
    cerr << "Line(" << LINE_NUM << ") ";
  stringstream ss;
  ss << H;
  cerr << args[idx] << " = " << ss.str();
  debug_out(args, idx + 1, LINE_NUM, T...);
}
int32_t main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    string a, b;
    cin >> a >> b;
    vector<long long> res;
    for (long long i = 0; i < a.size(); i++) {
      if (a[i] == '1') {
        long long j = i;
        while ((j < a.size()) && (a[j] == '1')) {
          j++;
        }
        if (i == 0) {
          res.push_back(j);
        } else {
          res.push_back(i);
          res.push_back(j);
        }
        i = j - 1;
      }
    }
    vector<long long> tmp;
    for (long long i = 0; i < b.size(); i++) {
      if (b[i] == '1') {
        long long j = i;
        while ((j < b.size()) && (b[j] == '1')) {
          j++;
        }
        if (i == 0) {
          tmp.push_back(j);
        } else {
          tmp.push_back(i);
          tmp.push_back(j);
        }
        i = j - 1;
      }
    }
    reverse(tmp.begin(), tmp.end());
    for (long long i = 0; i < tmp.size(); i++) res.push_back(tmp[i]);
    cout << res.size() << " ";
    for (long long i = 0; i < res.size(); i++) cout << res[i] << " ";
    cout << "\n";
  }
  return 0;
}