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;
long long int mod = 1e9 + 7;
using namespace std::chrono;
vector<vector<pair<long long int, long long int>>> vec;
vector<long long int> vis;
long long int ans = 0;
void dfs(long long int i, long long int par, long long int sum) {
  if (vis[i]) return;
  vis[i] = 1;
  for (auto x : vec[i]) {
    if (!vis[x.first]) {
      ans = max(ans, sum + x.second);
      dfs(x.first, i, sum + x.second);
    }
  }
}
int main() {
  long long int q;
  cin >> q;
  while (q--) {
    long long int n;
    cin >> n;
    string a, b;
    cin >> a >> b;
    vector<long long int> ans;
    long long int l = 0, r = n - 1, ctr = 0;
    long long int flag = 0;
    for (int i = n - 1; i >= 0; i--) {
      if (flag == 0) {
        if (a[l] == b[i]) {
          ans.push_back(1);
          ans.push_back(i + 1);
          flag = 1;
          l++;
        } else {
          ans.push_back(i + 1);
          flag = 1;
          l++;
        }
      } else {
        if (a[r] != b[i]) {
          ans.push_back(1);
          ans.push_back(i + 1);
          flag = 0;
          r--;
        } else {
          ans.push_back(i + 1);
          flag = 0;
          r--;
        }
      }
    }
    cout << ans.size() << " ";
    for (int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
    cout << endl;
  }
  return 0;
}