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;
template <typename T>
T binpow(T a, T n, T m) {
  T r = 1;
  while (n) {
    if (n & 1) r = (r * a) % m;
    a = (a * a) % m;
    n /= 2;
  }
  return r;
}
long long isprime(long long n) {
  if (n == 0 || n == 1) return n;
  if (n % 2 == 0) return (long long)2;
  if (n % 3 == 0)
    return (long long)3;
  else {
    for (long long i = 5; i * i <= n; i += 6) {
      if (n % i == 0) return i;
      if (n % (i + 2) == 0) return (i + 2);
    }
    return n;
  }
}
const long long N = 2e5 + 5;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  srand(time(NULL));
  ;
  long long l, r, m, n, i, j, x, y, t;
  cin >> t;
  while (t--) {
    cin >> n;
    string a, b;
    cin >> a >> b;
    string s = a;
    vector<long long> ans;
    for (i = n - 1; i >= 0; --i) {
      if (s[i] == b[i]) continue;
      if (s[0] != b[i]) {
        for (j = 0; j <= i; ++j) {
          if (s[j] == '0')
            s[j] = '1';
          else
            s[j] = '0';
        }
        reverse(s.begin(), s.begin() + i + 1);
        ans.push_back(i + 1);
      } else {
        if (s[0] == '0')
          s[0] = '1';
        else
          s[0] = '0';
        ans.push_back(1);
        for (j = 0; j <= i; ++j) {
          if (s[j] == '0')
            s[j] = '1';
          else
            s[j] = '0';
        }
        reverse(s.begin(), s.begin() + i + 1);
        ans.push_back(i + 1);
      }
    }
    cout << ans.size() << " ";
    for (long long(j) = (0); (j) < (ans.size()); ++(j)) cout << ans[j] << " ";
    cout << ""
         << "\n";
    ;
    ;
  }
}