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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    string s;
    string target;
    cin >> s >> target;
    vector<int> steps;
    for (int i = 0, j = n - 1; i < s.length() and j >= 0;) {
      if (i == j) {
        if (s[j] == target[j]) {
          j--;
          continue;
        } else {
          steps.push_back(1);
          j--;
          if (s[0] == '1') {
            s[0] = '0';
          } else {
            s[0] = '1';
          }
          continue;
        }
      }
      char a = s[i];
      char b = target[j];
      int aint = a - '0';
      int bint = b - '0';
      if (target[j] == s[j]) {
        j--;
        continue;
      } else {
        if (aint ^ bint > 0) {
          steps.push_back(j + 1);
          reverse(s.begin() + i, s.begin() + j + 1);
          for (int k = 0; k <= j; k++) {
            if (s[k] == '1') {
              s[k] = '0';
            } else {
              s[k] = '1';
            }
          }
        } else {
          steps.push_back(1);
          steps.push_back(j + 1);
          if (s[i] == '1') {
            s[i] = '0';
          } else {
            s[i] = '1';
          }
          reverse(s.begin(), s.begin() + j + 1);
          for (int k = 0; k <= j; k++) {
            if (s[k] == '1') {
              s[k] = '0';
            } else {
              s[k] = '1';
            }
          }
        }
      }
      j--;
    }
    cout << steps.size() << " ";
    if (steps.size() == 0) {
      continue;
    }
    for (int i = 0; i < steps.size(); i++) {
      cout << steps[i] << " ";
    }
    cout << endl;
  }
}
