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;
const long long MOD = 1e9 + 7;
const long long MAXN = 1e6 + 7;
const long long INF = LONG_LONG_MAX;
long long n;
string a, b;
char OR(char c) {
  if (c == '0')
    return '1';
  else
    return '0';
}
void change(long long x) {
  string t = a.substr(0, x + 1);
  reverse((t).begin(), (t).end());
  for (long long i = 0; i < (x + 1); i++) a[i] = t[i];
  for (long long i = 0; i < (x + 1); i++) a[i] = OR(a[i]);
}
void solve() {
  cin >> n;
  cin >> a >> b;
  long long till = 0, back = 0;
  long long i = n - 1, front = 0;
  vector<long long> moves;
  while (i >= 0) {
    if (b[i] == a[0]) {
      moves.push_back(1);
      a[0] = OR(b[i]);
    }
    moves.push_back(i + 1);
    change(i);
    i--;
    front++;
  }
  cout << moves.size() << ' ';
  for (auto x : moves) cout << x << " ";
  cout << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) solve();
  cerr << "[ Time : " << (float)clock() / CLOCKS_PER_SEC << " secs ]" << endl;
}