#include <bits/stdc++.h>
using namespace std;
template <typename T1, typename T2>
string print_iterable(T1 begin_iter, T2 end_iter, int counter) {
  bool done_something = false;
  stringstream res;
  res << "[";
  for (; begin_iter != end_iter and counter; ++begin_iter) {
    done_something = true;
    counter--;
    res << *begin_iter << ", ";
  }
  string str = res.str();
  if (done_something) {
    str.pop_back();
    str.pop_back();
  }
  str += "]";
  return str;
}
vector<int> SortIndex(int size, std::function<bool(int, int)> compare) {
  vector<int> ord(size);
  for (int i = 0; i < size; i++) ord[i] = i;
  sort(ord.begin(), ord.end(), compare);
  return ord;
}
template <typename T>
bool MinPlace(T& a, const T& b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <typename T>
bool MaxPlace(T& a, const T& b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <typename S, typename T>
ostream& operator<<(ostream& out, const pair<S, T>& p) {
  out << "{" << p.first << ", " << p.second << "}";
  return out;
}
template <typename T>
ostream& operator<<(ostream& out, const vector<T>& v) {
  out << "[";
  for (int i = 0; i < (int)v.size(); i++) {
    out << v[i];
    if (i != (int)v.size() - 1) out << ", ";
  }
  out << "]";
  return out;
}
template <class TH>
void _dbg(const char* name, TH val) {
  clog << name << ": " << val << endl;
}
template <class TH, class... TA>
void _dbg(const char* names, TH curr_val, TA... vals) {
  while (*names != ',') clog << *names++;
  clog << ": " << curr_val << ", ";
  _dbg(names + 1, vals...);
}
void check(string A, string B, vector<int> moves) {
  for (int x : moves) {
    for (int i = 0; i <= x; i++) A[i] = (A[i] == '0') ? '1' : '0';
    for (int i = 0; i < x - i; i++) swap(A[i], A[x - i]);
  }
  assert(A == B);
}
void solve() {
  int N;
  cin >> N;
  string A;
  string B;
  cin >> A >> B;
  int cnt = 0;
  int sum = 0;
  auto get_val = [&](int id) {
    if (cnt % 2 == 0) return A[sum + id];
    return (A[sum - id] == '0') ? '1' : '0';
  };
  vector<int> moves;
  for (int i = N - 1; i >= 0; i--) {
    if (get_val(i) == B[i]) continue;
    if (B[i] != get_val(0))
      moves.push_back(i);
    else {
      moves.push_back(0);
      moves.push_back(i);
    }
    cnt++;
    if (cnt % 2)
      sum = sum + i;
    else
      sum = sum - i;
  }
  cout << moves.size() << " ";
  for (int x : moves) cout << x + 1 << " ";
  cout << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  for (int t = 0; t < T; t++) solve();
}

### Prompt

Your challenge is to write 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;
template <typename T1, typename T2>
string print_iterable(T1 begin_iter, T2 end_iter, int counter) {
  bool done_something = false;
  stringstream res;
  res << "[";
  for (; begin_iter != end_iter and counter; ++begin_iter) {
    done_something = true;
    counter--;
    res << *begin_iter << ", ";
  }
  string str = res.str();
  if (done_something) {
    str.pop_back();
    str.pop_back();
  }
  str += "]";
  return str;
}
vector<int> SortIndex(int size, std::function<bool(int, int)> compare) {
  vector<int> ord(size);
  for (int i = 0; i < size; i++) ord[i] = i;
  sort(ord.begin(), ord.end(), compare);
  return ord;
}
template <typename T>
bool MinPlace(T& a, const T& b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <typename T>
bool MaxPlace(T& a, const T& b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <typename S, typename T>
ostream& operator<<(ostream& out, const pair<S, T>& p) {
  out << "{" << p.first << ", " << p.second << "}";
  return out;
}
template <typename T>
ostream& operator<<(ostream& out, const vector<T>& v) {
  out << "[";
  for (int i = 0; i < (int)v.size(); i++) {
    out << v[i];
    if (i != (int)v.size() - 1) out << ", ";
  }
  out << "]";
  return out;
}
template <class TH>
void _dbg(const char* name, TH val) {
  clog << name << ": " << val << endl;
}
template <class TH, class... TA>
void _dbg(const char* names, TH curr_val, TA... vals) {
  while (*names != ',') clog << *names++;
  clog << ": " << curr_val << ", ";
  _dbg(names + 1, vals...);
}
void check(string A, string B, vector<int> moves) {
  for (int x : moves) {
    for (int i = 0; i <= x; i++) A[i] = (A[i] == '0') ? '1' : '0';
    for (int i = 0; i < x - i; i++) swap(A[i], A[x - i]);
  }
  assert(A == B);
}
void solve() {
  int N;
  cin >> N;
  string A;
  string B;
  cin >> A >> B;
  int cnt = 0;
  int sum = 0;
  auto get_val = [&](int id) {
    if (cnt % 2 == 0) return A[sum + id];
    return (A[sum - id] == '0') ? '1' : '0';
  };
  vector<int> moves;
  for (int i = N - 1; i >= 0; i--) {
    if (get_val(i) == B[i]) continue;
    if (B[i] != get_val(0))
      moves.push_back(i);
    else {
      moves.push_back(0);
      moves.push_back(i);
    }
    cnt++;
    if (cnt % 2)
      sum = sum + i;
    else
      sum = sum - i;
  }
  cout << moves.size() << " ";
  for (int x : moves) cout << x + 1 << " ";
  cout << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  for (int t = 0; t < T; t++) solve();
}

```