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.
#include <bits/stdc++.h>
using namespace std;
const long long int inf = 9e18;
const long double pi = 2 * acos(0.0);
vector<long long int> ans, pos;
string str1, str2;
void flip(long long int len, long long int k) {
  long long int off = pos[k];
  if (len == 1) {
    if (k % 2 == 0) {
      if (str1[off] != str2[len - 1]) ans.push_back(0);
      return;
    } else {
      if (str1[off] == str2[len - 1]) ans.push_back(0);
      return;
    }
  }
  if (off == 0) {
    if (str1[off] != str2[len - 1]) {
      ans.push_back(len - 1);
      flip(len - 1, k + 1);
    } else {
      ans.push_back(0);
      ans.push_back(len - 1);
      flip(len - 1, k + 1);
    }
  } else {
    if (k % 2 == 0) {
      if (str1[off] != str2[len - 1]) {
        ans.push_back(len - 1);
        flip(len - 1, k + 1);
      } else {
        ans.push_back(0);
        ans.push_back(len - 1);
        flip(len - 1, k + 1);
      }
    } else {
      if (str1[off] != str2[len - 1]) {
        ans.push_back(0);
        ans.push_back(len - 1);
        flip(len - 1, k + 1);
      } else {
        ans.push_back(len - 1);
        flip(len - 1, k + 1);
      }
    }
    return;
  }
}
void solve() {
  long long int n;
  cin >> n;
  cin >> str1 >> str2;
  long long int l = 0, r = n - 1;
  while (l <= r) {
    pos.push_back(l);
    if (l != r) pos.push_back(r);
    l++;
    r--;
  }
  flip(n, 0);
  cout << (int)((ans).size()) << " ";
  for (auto it : ans) cout << it + 1 << " ";
  cout << "\n";
  while (!pos.empty()) pos.pop_back();
  while (!ans.empty()) ans.pop_back();
  ans.shrink_to_fit();
  pos.shrink_to_fit();
  return;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int tc = 1;
  cin >> tc;
  while (tc--) {
    solve();
  }
}
