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;
vector<vector<int> > edges((500) + 9);
vector<vector<int> > edges1((500) + 9);
vector<vector<int> > languages((500) + 9);
vector<set<int> > edges2((500) + 9);
vector<int> distances((500) + 9, -1);
vector<int> dfsvec;
vector<int> leafs;
bool visited2[(500) + 9];
long gcd(long a, long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
int lcm(int a, int b) { return (a * b) / gcd(a, b); }
void addedges(int v, int u) {
  edges[v].push_back(u);
  edges[u].push_back(v);
}
int count1;
void dfs(int node) {
  visited2[node] = true;
  for (auto x : edges2[node])
    if (!visited2[x]) {
      dfs(x);
    }
}
queue<int> path();
void bfs(int node) {
  queue<int> q;
  distances[node] = 0;
  q.push(node);
  while (!q.empty()) {
    int src = q.front();
    q.pop();
    int cost = distances[src];
    for (auto child : edges[src]) {
      if (distances[child] == -1) {
        q.push(child);
        distances[child] = cost + 1;
      }
    }
  }
}
int graph_height(int node) {
  int h = 0;
  for (int adj : edges[node]) {
    h = max(h, 1 + graph_height(adj));
  }
  return h;
}
int mod(int a, int b) { return (((a % b) + b) % b); }
namespace patch {
template <typename T>
std::string to_string(const T& n) {
  std::ostringstream stm;
  stm << n;
  return stm.str();
}
}  // namespace patch
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    string a, b;
    cin >> a >> b;
    vector<int> count;
    int cont = 0;
    vector<int> a1, b1;
    a += '0';
    b += '0';
    for (int i = 1; i <= n; i++) {
      if (a[i] != a[i - 1]) a1.push_back(i);
      if (b[i] != b[i - 1]) b1.push_back(i);
    }
    a1.insert(a1.end(), b1.rbegin(), b1.rend());
    cout << a1.size() << " ";
    for (auto x : a1) cout << x << " ";
    cout << '\n';
  }
}