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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long int mod = 1000000007;
const long long int N = 300001;
const long long int M = log2(N);
struct point {
  long long int x, y, i;
};
void debug(stringstream &s) {}
template <typename T, typename... Args>
void debug(stringstream &s, T a, Args... args) {
  string word;
  s >> word;
  cerr << word << " = " << a << endl;
  debug(s, args...);
}
long long int fast_exp(long long int x, long long int n) {
  if (n == 0) return 1;
  if (n & 1)
    return fast_exp(x, n - 1) * x % mod;
  else {
    long long int ans = fast_exp(x, n / 2);
    return ans * ans % mod;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int _ = 1;
  cin >> _;
  int test = 0;
  while (_--) {
    test++;
    long long int n, m, x, y = 0, k, ans = 0;
    cin >> n;
    string a, b;
    cin >> a >> b;
    vector<long long int> sol;
    x = 0, y = n - 1;
    bool state = 0;
    for (int i = n - 1; i > 0; i--) {
      if (a[i] != b[i]) {
        if (a[0] == b[i]) {
          sol.push_back(1);
          sol.push_back(i + 1);
          for (int j = 1; j <= i; j++) {
            if (a[j] == '1')
              a[j] = '0';
            else
              a[j] = '1';
          }
          for (int j = 0; j <= i / 2; j++) {
            if (j <= i - j) swap(a[j], a[i - j]);
          }
        } else {
          sol.push_back(i + 1);
          for (int j = 0; j <= i; j++) {
            if (a[j] == '1')
              a[j] = '0';
            else
              a[j] = '1';
          }
          for (int j = 0; j <= i / 2; j++) {
            if (j <= i - j) swap(a[j], a[i - j]);
          }
        }
        state = state ^ 1;
      }
    }
    if (a[0] != b[0]) sol.push_back(1);
    cout << sol.size() << "\n";
    for (auto i : sol) cout << i << " ";
    cout << "\n";
  }
  return 0;
}