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 int MAX = 1e9;
const int MOD = 1e9 + 7;
const long long INF = 1e18;
const long double PI = 3.14159265358979323846;
template <typename T>
void read(vector<T>& a) {
  for (int i = 0; i < a.size(); ++i) cin >> a[i];
}
template <typename T>
void print(vector<T>& a) {
  for (int i = 0; i < a.size(); ++i) cout << a[i] << " ";
  cout << endl;
}
template <typename T>
void print2(vector<vector<T> >& a) {
  for (int i = 0; i < a.size(); ++i) {
    for (int j = 0; j < a[i].size(); ++j) {
      cout << a[i][j] << " ";
    }
    cout << endl;
  }
}
template <typename T>
void read2(vector<vector<T> >& a) {
  for (int i = 0; i < a.size(); ++i) {
    for (int j = 0; j < a[i].size(); ++j) {
      cin >> a[i][j];
    }
  }
}
void write() {
  int n;
  cin >> n;
  string s, t;
  cin >> s >> t;
  vector<int> ans;
  for (int i = n - 1; i >= 0; --i) {
    if (t[i] == s[0]) {
      if (i == 0) continue;
      ans.push_back(1);
      if (s[0] == '0')
        s[0] = '1';
      else
        s[0] = '0';
    }
    ans.push_back(i + 1);
    string tmp = "";
    for (int j = 0; j <= i; j++) {
      if (s[j] == '0')
        tmp += "1";
      else
        tmp += "0";
    }
    reverse((tmp).begin(), (tmp).end());
    for (int j = 0; j <= i; j++) {
      s[j] = tmp[j];
    }
  }
  cout << ans.size() << " ";
  print(ans);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  t = 1;
  cin >> t;
  for (int T = 0; T < t; ++T) write();
  return 0;
}