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;
signed main() {
  int test = 1;
  cin >> test;
  while (test--) {
    int n;
    cin >> n;
    char a[n], b[n];
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int i = 0; i < n; i++) cin >> b[i];
    vector<int> ans;
    int cnt = 0;
    int xo = n;
    while (n--) {
      if (n == 0) {
        if (a[0] == b[0])
          break;
        else {
          ans.push_back(1);
          break;
        }
      }
      if (a[0] == b[n]) {
        ans.push_back(1);
        if (a[0] == '1')
          a[0] = '0';
        else
          a[0] = '1';
      }
      cnt++;
      ans.push_back(n + 1);
      char x;
      if (cnt % 2) {
        x = a[xo - 1 - cnt / 2];
      } else {
        x = a[cnt / 2];
        x = (x == '0' ? '1' : '0');
      }
      a[0] = (x == '0' ? '1' : '0');
    }
    cout << ans.size();
    for (auto i : ans) {
      cout << " " << i;
    }
    cout << endl;
  }
}
