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 inp, Inp;
long long i1, i2, i3, i4, i5, i6, i7, i8;
long long MOD = 998244353, MOD1 = (long long)1e9 + 7, MOD2 = (long long)1e9 + 9;
long long MOD3 = (long long)1e9 + 21;
double eps = 1e-9;
long long TMP =
    (chrono::high_resolution_clock::now().time_since_epoch().count() % MOD +
     MOD) %
    MOD;
auto STTMP = chrono::steady_clock::now();
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  std::cout.precision(20);
  srand((unsigned int)TMP);
  long long T;
  cin >> T;
  for (int t = (0); t < (T); t++) {
    long long N;
    cin >> N;
    string A;
    cin >> A;
    string B;
    cin >> B;
    long long pos = N - 1;
    long long abeg = 0, aend = N - 1;
    bool flipped = false;
    vector<long long> sol;
    while (pos >= 1) {
      if (A.at(abeg) == B.at(pos)) {
        sol.push_back(1);
      }
      sol.push_back(pos + 1);
      pos--;
      if (A.at(aend) != B.at(pos)) {
        sol.push_back(1);
      }
      sol.push_back(pos + 1);
      abeg++;
      aend--;
      pos--;
    }
    if (pos == 0) {
      if (A.at(abeg) != B.at(pos)) {
        sol.push_back(1);
      }
    }
    cout << sol.size() << " ";
    for (auto it : sol) {
      cout << it << " ";
    }
    cout << "\n";
  }
  return 0;
}