#include <bits/stdc++.h>
using namespace std;
template <typename Arg1>
void prn(Arg1&& arg1) {
  cout << arg1 << "\n";
}
template <typename Arg1, typename... Args>
void prn(Arg1&& arg1, Args&&... args) {
  cout << arg1 << " ";
  prn(args...);
}
template <typename Arg1>
void prs(Arg1&& arg1) {
  cout << arg1 << " ";
}
template <typename Arg1, typename... Args>
void prs(Arg1&& arg1, Args&&... args) {
  cout << arg1 << " ";
  prs(args...);
}
template <typename Arg1>
void read(Arg1&& arg1) {
  cin >> arg1;
}
template <typename Arg1, typename... Args>
void read(Arg1&& arg1, Args&&... args) {
  cin >> arg1;
  read(args...);
}
string a, b;
int n;
void solve() {
  read(n, a, b);
  int st = 0, en = n - 1;
  bool tog = 0;
  vector<int> ans;
  for (auto& i : a) i -= '0';
  for (auto& i : b) i -= '0';
  for (int pos = n - 1; pos >= 0; --pos) {
    if (st < en) {
      int cur = a[st] ^ tog;
      if (cur == b[pos]) {
        ans.push_back(1);
      }
      ans.push_back(pos + 1);
      swap(st, en);
      ++en;
      tog ^= 1;
    } else {
      int cur = a[st] ^ tog;
      if (cur == b[pos]) {
        ans.push_back(1);
      }
      ans.push_back(pos + 1);
      swap(st, en);
      --en;
      tog ^= 1;
    }
  }
  prs(ans.size());
  for (auto i : ans) prs(i);
  prn("");
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  read(t);
  for (int tc = 1; tc <= t; ++tc) {
    solve();
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

This is the hard 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 2n 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≤ 10^5) — 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 10^5.

Output

For each test case, output an integer k (0≤ k≤ 2n), 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <typename Arg1>
void prn(Arg1&& arg1) {
  cout << arg1 << "\n";
}
template <typename Arg1, typename... Args>
void prn(Arg1&& arg1, Args&&... args) {
  cout << arg1 << " ";
  prn(args...);
}
template <typename Arg1>
void prs(Arg1&& arg1) {
  cout << arg1 << " ";
}
template <typename Arg1, typename... Args>
void prs(Arg1&& arg1, Args&&... args) {
  cout << arg1 << " ";
  prs(args...);
}
template <typename Arg1>
void read(Arg1&& arg1) {
  cin >> arg1;
}
template <typename Arg1, typename... Args>
void read(Arg1&& arg1, Args&&... args) {
  cin >> arg1;
  read(args...);
}
string a, b;
int n;
void solve() {
  read(n, a, b);
  int st = 0, en = n - 1;
  bool tog = 0;
  vector<int> ans;
  for (auto& i : a) i -= '0';
  for (auto& i : b) i -= '0';
  for (int pos = n - 1; pos >= 0; --pos) {
    if (st < en) {
      int cur = a[st] ^ tog;
      if (cur == b[pos]) {
        ans.push_back(1);
      }
      ans.push_back(pos + 1);
      swap(st, en);
      ++en;
      tog ^= 1;
    } else {
      int cur = a[st] ^ tog;
      if (cur == b[pos]) {
        ans.push_back(1);
      }
      ans.push_back(pos + 1);
      swap(st, en);
      --en;
      tog ^= 1;
    }
  }
  prs(ans.size());
  for (auto i : ans) prs(i);
  prn("");
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  read(t);
  for (int tc = 1; tc <= t; ++tc) {
    solve();
  }
  return 0;
}

```