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>
const double PI = 3.141592653589793238460;
using namespace std;
long long pows(long long a, long long n, long long m) {
  long long res = 1;
  while (n) {
    if (n % 2 != 0) {
      res = (res * a) % m;
      n--;
    } else {
      a = (a * a) % m;
      n = n / 2;
    }
  }
  return res % m;
}
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
bool isprime(long long n) {
  if (n == 1) {
    return false;
  }
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      return false;
    }
  }
  return true;
}
bool istrue(string s) {
  int i = 0;
  int j = s.size() - 1;
  while (i < j) {
    if (s[i] == s[j]) {
      i++;
      j--;
    } else {
      return false;
    }
  }
  return true;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    string s, p;
    cin >> s >> p;
    int cnt = 0;
    vector<int> vi;
    cnt = 0;
    for (int i = n - 1; i >= 0; i--) {
      if (s[i] != p[i]) {
        if (s[0] != p[i]) {
          vi.push_back(i + 1);
        } else {
          vi.push_back(1);
          vi.push_back(i + 1);
        }
        for (int j = 0; j <= i; j++) {
          if (s[j] == '0') {
            s[j] = '1';
          } else {
            s[j] = '0';
          }
        }
        reverse(s.begin(), s.end() - cnt);
      }
      cnt++;
    }
    cout << vi.size() << " ";
    for (int res : vi) {
      cout << res << " ";
    }
    cout << endl;
  }
}