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 double pi = acos(-1.0);
const double pii = 2 * pi;
const double eps = 1e-6;
const double inf = 1e15;
char flip(char x) {
  if (x == '0')
    return '1';
  else
    return '0';
}
void solve() {
  long long int n;
  cin >> n;
  ;
  string a, b;
  cin >> a >> b;
  long long int count = 0;
  vector<long long int> ans;
  long long int f = 0;
  long long int j = 0;
  for (long long int i = n - 1; i >= 0; i--) {
    if (f == 0) {
      int x = (a[j] + count - '0') % 2;
      char xx = '0' + x;
      if (b[i] == xx) {
        ans.push_back(1);
        a[j] = flip(xx);
      }
      ans.push_back(i + 1);
      f ^= 1;
      count++;
    } else {
      int x = (a[n - 1 - j] + count - '0') % 2;
      char xx = '0' + x;
      if (b[i] == xx) {
        ans.push_back(1);
        a[n - 1 - j] = flip(xx);
      }
      ans.push_back(i + 1);
      f ^= 1;
      j++;
      count++;
    }
  }
  cout << ans.size() << " ";
  for (auto i : ans) cout << i << " ";
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  cin >> t;
  for (long long int i = 1; i <= t; i++) {
    solve();
  }
  cerr << "Time : " << (double)clock() / (double)CLOCKS_PER_SEC << "s\n";
  return 0;
}