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;
template <typename T>
void sorted(vector<long long int> &arr) {
  sort(arr.begin(), arr.end());
}
void coutarr(vector<long long int> arr) {
  long long int n = arr.size();
  for (long long int i = 0; i < n; i++) cout << arr[i] << " ";
  cout << '\n';
}
void coutarr2(vector<vector<long long int>> arr) {
  long long int n = arr.size();
  for (long long int i = 0; i < n; i++) {
    for (long long int j = 0; j < arr[i].size(); j++) cout << arr[i][j] << " ";
    cout << '\n';
  }
}
void couta(long long int x) { cout << to_string(x) << " " << x << '\n'; }
void coutpair(vector<pair<long long int, long long int>> arr) {
  long long int n = arr.size();
  for (long long int i = 0; i < n; i++) cout << arr[i].first << " ";
  cout << '\n';
  for (long long int i = 0; i < n; i++) cout << arr[i].second << " ";
  cout << '\n';
}
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
string revbit(string &s, long long int k) {
  for (long long int i = 0; i <= k; i++) {
    if (s[i] == '1')
      s[i] = '0';
    else
      s[i] = '1';
  }
  for (long long int i = 0; i < (k + 1) / 2; i++) swap(s[i], s[k - i]);
  return s;
}
void solve() {
  long long int n;
  cin >> n;
  string a, b;
  cin >> a >> b;
  long long int ct = n - 1;
  vector<long long int> ans;
  while (ct >= 0) {
    if (b[ct] != a[0]) {
      a = revbit(a, ct);
      ans.push_back(ct + 1);
    } else {
      if (a[0] == '1')
        a[0] = '0';
      else
        a[0] = '1';
      ans.push_back(1);
    }
    while (a[ct] == b[ct]) ct--;
  }
  cout << ans.size() << " ";
  coutarr(ans);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int u = 1;
  cin >> u;
  while (u--) {
    solve();
  }
  return 0;
}