#include <bits/stdc++.h>
using namespace std;
void tokenize(string str, const char delim, vector<string>& res) {
  stringstream p(str);
  string token;
  while (getline(p, token, delim)) {
    res.push_back(token);
  }
  return;
}
int N;
string str1, str2;
void test_case() {
  cin >> N >> str1 >> str2;
  vector<int> ans;
  int end = N - 1, start = 0;
  int ptr = 0;
  int flips = 0;
  for (int i = N - 1; i >= 0; --i) {
    if (flips % 2 != 0) str1[ptr] ^= 1;
    if (str2[i] == str1[ptr]) ans.push_back(0);
    ans.push_back(i);
    ++flips;
    if (ptr == start)
      ++start, ptr = end;
    else
      --end, ptr = start;
  }
  cout << ((long long)ans.size()) << ' ';
  for (auto el : ans) cout << el + 1 << ' ';
  cout << endl;
}
int main() {
  ios_base ::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
  ;
  int tt;
  cin >> tt;
  while (tt--) {
    test_case();
  }
  return 0;
}

### Prompt

Your task is to create a cpp solution to the following problem:

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;
void tokenize(string str, const char delim, vector<string>& res) {
  stringstream p(str);
  string token;
  while (getline(p, token, delim)) {
    res.push_back(token);
  }
  return;
}
int N;
string str1, str2;
void test_case() {
  cin >> N >> str1 >> str2;
  vector<int> ans;
  int end = N - 1, start = 0;
  int ptr = 0;
  int flips = 0;
  for (int i = N - 1; i >= 0; --i) {
    if (flips % 2 != 0) str1[ptr] ^= 1;
    if (str2[i] == str1[ptr]) ans.push_back(0);
    ans.push_back(i);
    ++flips;
    if (ptr == start)
      ++start, ptr = end;
    else
      --end, ptr = start;
  }
  cout << ((long long)ans.size()) << ' ';
  for (auto el : ans) cout << el + 1 << ' ';
  cout << endl;
}
int main() {
  ios_base ::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
  ;
  int tt;
  cin >> tt;
  while (tt--) {
    test_case();
  }
  return 0;
}

```