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;
void DBG() { cerr << "]" << endl; }
template <class H, class... T>
void DBG(H h, T... t) {
  cerr << to_string(h);
  if (sizeof...(t)) cerr << ", ";
  DBG(t...);
}
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int Q;
  cin >> Q;
  while (Q--) {
    int N;
    cin >> N;
    string S1, S2;
    cin >> S1 >> S2;
    vector<int> v;
    int l = S1[0] - '0', r;
    int cnt = 0;
    int lb = 0, rb = N;
    for (int i = N - 1; i >= 0; --i) {
      if (lb < rb)
        --rb;
      else
        ++rb;
      l = S1[lb] - '0';
      r = S1[rb] - '0';
      if (cnt & 1) r = r ^ 1;
      if (cnt & 1) l = l ^ 1;
      if (r == S2[i] - '0') continue;
      if (i == 0)
        v.push_back(1);
      else {
        if (l == r) {
          v.push_back(i + 1);
        } else {
          v.push_back(1);
          v.push_back(i + 1);
        }
        ++cnt;
        swap(lb, rb);
      }
    }
    cout << ((int)(v).size()) << ' ';
    for (auto x : v) cout << x << ' ';
    cout << "\n";
  }
}