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;
void solve() {
  long long int n;
  cin >> n;
  string s, s1;
  cin >> s >> s1;
  deque<char> q, q1;
  for (auto it : s) {
    q.push_back(it);
  }
  for (auto it : s1) {
    q1.push_back(it);
  }
  vector<long long int> ans;
  int check = 0;
  long long int curr = n;
  while (!q1.empty()) {
    if (check == 0) {
      if (q.back() == q1.back()) {
        q.pop_back();
        q1.pop_back();
      } else {
        if (q.front() != q1.back()) {
          ans.push_back(curr);
          check = 1;
          q.pop_front();
          q1.pop_back();
        } else {
          check = 1;
          ans.push_back(1);
          ans.push_back(curr);
          q.pop_front();
          q1.pop_back();
        }
      }
    } else {
      if (q.front() != q1.back()) {
        q.pop_front();
        q1.pop_back();
      } else {
        if (q.back() == q1.back()) {
          ans.push_back(curr);
          check = 0;
          q.pop_back();
          q1.pop_back();
        } else {
          check = 0;
          ans.push_back(1);
          ans.push_back(curr);
          q.pop_back();
          q1.pop_back();
        }
      }
    }
    curr--;
  }
  cout << ans.size() << " ";
  for (auto it : ans) {
    cout << it << " ";
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t = 1;
  cin >> t;
  for (long long int i = 0; i < t; i++) {
    solve();
  }
  return 0;
}