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 ll = long long;
using ull = unsigned long long;
void op(int len, std::string& a) {
  std::string amod(a);
  for (auto i = 0; i < len; i++) {
    a[i] = amod[len - 1 - i] == '0' ? '1' : '0';
  }
}
void solve(int n, std::string& a, std::string& b, std::vector<int>& r) {
  if (n == 0) {
    return;
  }
  if (a[n - 1] == b[n - 1]) {
    solve(n - 1, a, b, r);
    return;
  }
  if (a[0] != b[n - 1]) {
    op(n, a);
    r.push_back(n);
    solve(n - 1, a, b, r);
    return;
  } else {
    op(1, a);
    op(n, a);
    r.push_back(1);
    r.push_back(n);
    solve(n - 1, a, b, r);
    return;
  }
}
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(NULL);
  int t;
  std::cin >> t;
  while (t--) {
    int n;
    std::cin >> n;
    std::string a;
    std::string b;
    std::cin >> a >> b;
    std::vector<int> r;
    solve(n, a, b, r);
    int k = r.size();
    std::cout << k << ' ';
    for (auto i = 0; i < k; i++) {
      std::cout << r[i] << ' ';
    }
    std::cout << '\n';
  }
  return 0;
}