#include <bits/stdc++.h>
using namespace std;
void debug_out() { cerr << "\n"; }
template <typename H, typename... T>
void debug_out(H h, T... t) {
  cerr << " " << (h);
  debug_out(t...);
}
int n;
string a, b;
char flip(char ch) { return (char)('0' + ('1' - ch)); }
char ch(int i, int cnt) { return (char)('0' + ((a[i] - '0' + cnt) % 2)); }
void solve() {
  cin >> n >> a >> b;
  vector<int> res;
  int first = 0;
  int last = n - 1;
  int cnt = 0;
  for (int i = n - 1; i >= 0; i--) {
    if (ch(last, cnt) != b[i]) {
      if (ch(first, cnt) == b[i]) {
        res.push_back(1);
        a[first] = flip(a[first]);
      }
      if (i or ch(first, cnt) != b[i]) {
        res.push_back(i + 1);
        cnt++;
        swap(first, last);
      }
    }
    if (cnt % 2) {
      last++;
    } else {
      last--;
    }
  }
  cout << res.size();
  for (auto i : res) cout << " " << i;
  cout << "\n";
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Please create a solution in Cpp 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 debug_out() { cerr << "\n"; }
template <typename H, typename... T>
void debug_out(H h, T... t) {
  cerr << " " << (h);
  debug_out(t...);
}
int n;
string a, b;
char flip(char ch) { return (char)('0' + ('1' - ch)); }
char ch(int i, int cnt) { return (char)('0' + ((a[i] - '0' + cnt) % 2)); }
void solve() {
  cin >> n >> a >> b;
  vector<int> res;
  int first = 0;
  int last = n - 1;
  int cnt = 0;
  for (int i = n - 1; i >= 0; i--) {
    if (ch(last, cnt) != b[i]) {
      if (ch(first, cnt) == b[i]) {
        res.push_back(1);
        a[first] = flip(a[first]);
      }
      if (i or ch(first, cnt) != b[i]) {
        res.push_back(i + 1);
        cnt++;
        swap(first, last);
      }
    }
    if (cnt % 2) {
      last++;
    } else {
      last--;
    }
  }
  cout << res.size();
  for (auto i : res) cout << " " << i;
  cout << "\n";
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```