#include <bits/stdc++.h>
using namespace std;
void solve() {
  long long n;
  cin >> n;
  string s, t;
  cin >> s >> t;
  deque<char> Q((s).begin(), (s).end());
  bool f = 1;
  vector<long long> ans;
  for (long long i = n - 1; i >= 0; i--) {
    if (f) {
      if (Q.back() == t[i]) {
        Q.pop_back();
        continue;
      }
      if (Q.front() == t[i]) {
        ans.push_back(1);
      }
      ans.push_back(i + 1);
      Q.pop_front();
    } else {
      if (Q.front() != t[i]) {
        Q.pop_front();
        continue;
      }
      if (Q.back() != t[i]) {
        ans.push_back(1);
      }
      ans.push_back(i + 1);
      Q.pop_back();
    }
    f = 1 - f;
  }
  cout << ((long long)(ans).size());
  for (long long i = 0; i < ((long long)(ans).size()); i++) {
    cout << ' ' << ans[i];
  }
  cout << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

### Prompt

Generate 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 solve() {
  long long n;
  cin >> n;
  string s, t;
  cin >> s >> t;
  deque<char> Q((s).begin(), (s).end());
  bool f = 1;
  vector<long long> ans;
  for (long long i = n - 1; i >= 0; i--) {
    if (f) {
      if (Q.back() == t[i]) {
        Q.pop_back();
        continue;
      }
      if (Q.front() == t[i]) {
        ans.push_back(1);
      }
      ans.push_back(i + 1);
      Q.pop_front();
    } else {
      if (Q.front() != t[i]) {
        Q.pop_front();
        continue;
      }
      if (Q.back() != t[i]) {
        ans.push_back(1);
      }
      ans.push_back(i + 1);
      Q.pop_back();
    }
    f = 1 - f;
  }
  cout << ((long long)(ans).size());
  for (long long i = 0; i < ((long long)(ans).size()); i++) {
    cout << ' ' << ans[i];
  }
  cout << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

```