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 int PI = 3.1415926535;
const int inf = 1000111222;
const long long inf64 = 1000000111111222222;
const int mod = 1000000007;
const int N = 100007;
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    string sa, sb;
    cin >> sa >> sb;
    deque<int> Q;
    for (auto &x : sa) {
      Q.push_back(x - '0');
    }
    int check = 0;
    vector<int> answer, br;
    for (auto &x : sb) {
      br.emplace_back(x - '0');
    }
    for (int i = n - 1; i > 0; i--) {
      if (!check) {
        if (Q.back() != br[i]) {
          if (Q.front() == br[i]) {
            Q.front() ^= 1;
            answer.emplace_back(1);
          }
          check ^= 1;
          answer.emplace_back(i + 1);
          Q.pop_front();
        } else
          Q.pop_back();
      } else {
        if (Q.front() == br[i]) {
          if (Q.back() != br[i]) {
            Q.back() ^= 1;
            answer.emplace_back(1);
          }
          check ^= 1;
          answer.emplace_back(i + 1);
          Q.pop_back();
        } else
          Q.pop_front();
      }
    }
    assert((int)Q.size() == 1);
    if ((!check && Q.back() != br[0]) || (check && Q.front() == br[0])) {
      answer.emplace_back(1);
    }
    cout << (int)answer.size() << " ";
    for (auto x : answer) {
      cout << x << " ";
    }
    cout << "\n";
  }
  return 0;
}