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;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
void solve() {
  long long n;
  cin >> n;
  string a, b;
  cin >> a >> b;
  vector<long long> c(n);
  for (long long i = 0; i < n; i++) {
    c[i] = (b[i] - '0');
  }
  vector<long long> p;
  long long cnt = 0;
  deque<long long> q;
  for (long long i = 0; i < n; i++) {
    q.push_back(a[i] - '0');
  }
  for (long long i = n - 1; i >= 0; i--) {
    if (cnt & 1) {
      long long ft = q.back();
      long long lt = q.front();
      ft ^= 1;
      lt ^= 1;
      if (lt == c[i]) {
        q.pop_front();
        continue;
      }
      if (ft == lt) {
        p.push_back(i + 1);
        q.pop_back();
        cnt++;
      } else {
        p.push_back(1);
        p.push_back(i + 1);
        q.pop_back();
        cnt++;
      }
    } else {
      long long ft = q.front();
      long long lt = q.back();
      if (lt == c[i]) {
        q.pop_back();
        continue;
      }
      if (ft == lt) {
        p.push_back(i + 1);
        q.pop_front();
        cnt++;
      } else {
        p.push_back(1);
        p.push_back(i + 1);
        q.pop_front();
        cnt++;
      }
    }
  }
  cout << (long long)(p).size() << " ";
  for (auto &el : p) {
    cout << el << " ";
  }
  cout << "\n";
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}