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 maxn = 1e5 + 5;
int T, n;
vector<int> v1, v2, res;
string str1, str2;
void solve() {
  v1.clear();
  v2.clear();
  res.clear();
  for (int i = 0; i < int(n); i++) {
    v1.push_back(str1[i] - '0');
    v2.push_back(str2[i] - '0');
  }
  int round = 0, idx = n - 1, left = 0, right = n - 1;
  for (int i = int(n - 1); i >= 0; i--) {
    if (v1[i] != v2[i]) {
      break;
    }
    idx--;
    right--;
  }
  while (idx >= 0) {
    int cur = round & 1 ? right-- : left++;
    if (round++ & 1) {
      v1[cur] ^= 1;
    }
    if (v1[cur] == v2[idx]) {
      res.push_back(1);
    }
    res.push_back(idx + 1);
    idx--;
  }
  cout << res.size();
  for (int i = 0; i < int(res.size()); i++) {
    cout << " " << res[i];
  }
  cout << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> T;
  while (T--) {
    cin >> n >> str1 >> str2;
    solve();
  }
  return 0;
}