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>
int exp(int x, int y, int p) {
  int res = 1;
  while (y) {
    if (y % 2) res = (res * x % p) % p;
    x = (x * x) % p;
    y /= 2;
  }
  return res;
}
int expo(int x, int y) {
  int res = 1;
  while (y) {
    if (y % 2) res = (res * x % ((int)1e9 + 7)) % ((int)1e9 + 7);
    x = (x * x) % ((int)1e9 + 7);
    y /= 2;
  }
  return res;
}
long long add(long long a, long long b) {
  return (a % ((int)1e9 + 7) + b % ((int)1e9 + 7) + ((int)1e9 + 7)) %
         ((int)1e9 + 7);
}
int sub(int a, int b) {
  return (a % ((int)1e9 + 7) - b % ((int)1e9 + 7) + ((int)1e9 + 7)) %
         ((int)1e9 + 7);
}
long long mul(long long a, long long b) {
  return ((a % ((int)1e9 + 7)) * (b % ((int)1e9 + 7)) + ((int)1e9 + 7)) %
         ((int)1e9 + 7);
}
int inv(int x) { return expo(x, ((int)1e9 + 7) - 2); }
int sv[2000006] = {0};
using namespace std;
string change_string(string s, long long index) {
  string ans = s;
  for (int i = 0; i <= index; i++) {
    if (s[index - i] == '0')
      ans[i] = '1';
    else
      ans[i] = '0';
  }
  return ans;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    string s1, s2;
    cin >> s1 >> s2;
    vector<long long> ans;
    for (int i = n - 1; i >= 0; i--) {
      if (s1[i] == s2[i])
        continue;
      else {
        if (s1[i] == s1[0]) {
          ans.push_back(i + 1);
          s1 = change_string(s1, i);
        } else {
          ans.push_back(1);
          if (s1[0] == '0')
            s1[0] = '1';
          else
            s1[0] = '0';
          ans.push_back(i + 1);
          s1 = change_string(s1, i);
        }
      }
    }
    cout << (long long)ans.size() << " ";
    for (int i = 0; i < (long long)ans.size(); i++) cout << ans[i] << " ";
    cout << "\n";
  }
}