#include <bits/stdc++.h>
using namespace std;
using ull = uint64_t;
using ll = int64_t;
using PII = pair<int, int>;
using VI = vector<int>;
string to_string(string s) { return '"' + s + '"'; }
string to_string(const char* s) { return to_string((string)s); }
string to_string(bool b) { return (b ? "true" : "false"); }
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A>
string to_string(A v) {
  bool first = true;
  string res = "{";
  for (const auto& x : v) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(x);
  }
  res += "}";
  return res;
}
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
const ll md = 1000000007;
void solve() {
  int n;
  string aa, b;
  cin >> n >> aa >> b;
  string a = aa;
  VI ans;
  if (a[0] != b[n - 1]) {
    ans.push_back(n);
  } else {
    ans.push_back(1);
    ans.push_back(n);
  }
  int k = 1;
  char last = 1 - (a[n - 1] - '0') + '0';
  int l = 1, r = n - 1;
  for (int64_t i = (n - 2), _b = (0); i >= _b; --i) {
    if (k) {
      if (a[r] != b[i]) {
        ans.push_back(1);
        ans.push_back(i + 1);
      } else {
        ans.push_back(i + 1);
      }
      r--;
    } else {
      if (a[l] == b[i]) {
        ans.push_back(1);
        ans.push_back(i + 1);
      } else {
        ans.push_back(i + 1);
      }
      l++;
    }
    k = 1 - k;
  }
  cout << ((int)(ans).size()) << " ";
  for (int k : ans) {
    cout << k << " ";
  }
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
}

### 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;
using ull = uint64_t;
using ll = int64_t;
using PII = pair<int, int>;
using VI = vector<int>;
string to_string(string s) { return '"' + s + '"'; }
string to_string(const char* s) { return to_string((string)s); }
string to_string(bool b) { return (b ? "true" : "false"); }
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A>
string to_string(A v) {
  bool first = true;
  string res = "{";
  for (const auto& x : v) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(x);
  }
  res += "}";
  return res;
}
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
const ll md = 1000000007;
void solve() {
  int n;
  string aa, b;
  cin >> n >> aa >> b;
  string a = aa;
  VI ans;
  if (a[0] != b[n - 1]) {
    ans.push_back(n);
  } else {
    ans.push_back(1);
    ans.push_back(n);
  }
  int k = 1;
  char last = 1 - (a[n - 1] - '0') + '0';
  int l = 1, r = n - 1;
  for (int64_t i = (n - 2), _b = (0); i >= _b; --i) {
    if (k) {
      if (a[r] != b[i]) {
        ans.push_back(1);
        ans.push_back(i + 1);
      } else {
        ans.push_back(i + 1);
      }
      r--;
    } else {
      if (a[l] == b[i]) {
        ans.push_back(1);
        ans.push_back(i + 1);
      } else {
        ans.push_back(i + 1);
      }
      l++;
    }
    k = 1 - k;
  }
  cout << ((int)(ans).size()) << " ";
  for (int k : ans) {
    cout << k << " ";
  }
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
}

```