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() {
  cin.tie(nullptr)->sync_with_stdio(false);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    string s;
    string s2;
    cin >> s;
    cin >> s2;
    vector<long long> arr(n);
    vector<long long> arr2(n);
    for (long long i = 0; i < n; i++) {
      arr[i] = (long long)s[i] - (long long)'0';
      arr2[i] = (long long)s2[i] - (long long)'0';
    }
    long long left = 0;
    long long right = n - 1;
    long long add = 0;
    long long count = 0;
    vector<long long> ans;
    while (left < right) {
      if (add % 2 == 0) {
        if ((arr[right] + add) % 2 == arr2[n - count - 1]) {
          right--;
          count++;
          continue;
        }
        if ((arr[left] + add) % 2 == arr2[n - count - 1]) {
          ans.push_back(1);
          arr[left] = (arr[left] + 1) % 2;
        }
        ans.push_back(n - count);
        for (long long i = left; i < right; i++) {
          if (arr[right - (i - left)] != arr[i]) {
            right--;
            count++;
          } else {
            break;
          }
        }
        add++;
      } else {
        if ((arr[left] + add) % 2 == arr2[n - count - 1]) {
          left++;
          count++;
          continue;
        }
        if ((arr[right] + add) % 2 == arr2[n - count - 1]) {
          ans.push_back(1);
          arr[right] = (arr[right] + 1) % 2;
        }
        ans.push_back(n - count);
        for (long long i = right; i > left; i--) {
          if (arr[left + (right - i)] != arr[i]) {
            left++;
            count++;
          } else {
            break;
          }
        }
        add++;
      }
    }
    if ((arr[left] + add) % 2 != arr2[0]) ans.push_back(1);
    cout << ans.size() << " ";
    for (long long nxt : ans) {
      cout << nxt << " ";
    }
    cout << "\n";
  }
  return 0;
}
