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;
long long int pp(long long int a, long long int b) {
  if (b == 0) return 1;
  long long int t = pp(a, b / 2);
  if (b % 2 == 0)
    return t * t;
  else
    return t * t * a;
}
string ss(string s, long long int j) {
  for (long long int i = 0; i < j; i++) {
    if (s[i] == '0')
      s[i] = '1';
    else
      s[i] = '0';
  }
  long long int l = 0, r = j - 1;
  while (l < r) {
    swap(s[l], s[r]);
    l++;
    r--;
  }
  return s;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long int q;
  cin >> q;
  while (q--) {
    long long int i, j, k, n, m, x, y, l, r, c = 0;
    string s, s1, s2 = "";
    cin >> n;
    cin >> s >> s1;
    vector<long long int> v;
    for (i = n - 1; i > 0; i--) {
      if (s[i] == s1[i])
        continue;
      else {
        if ((s1[i] == '0' && s[i] == '1' && s[0] == '1') ||
            (s1[i] == '1' && s[i] == '0' && s[0] == '0')) {
          v.push_back(i + 1);
          s = ss(s, i + 1);
        } else if (s1[i] == '0' && s[i] == '1' && s[0] == '0') {
          v.push_back(1);
          s[0] = '1';
          v.push_back(i + 1);
          s = ss(s, i + 1);
        } else {
          v.push_back(1);
          s[0] = '0';
          v.push_back(i + 1);
          s = ss(s, i + 1);
        }
      }
    }
    if (s[0] != s1[0]) v.push_back(1);
    cout << v.size() << " ";
    for (i = 0; i < v.size(); i++) cout << v[i] << " ";
    cout << endl;
  }
}