#include <bits/stdc++.h>
using namespace std;
long long MX = INT_MAX;
long long MN = INT_MIN;
long long INF = 1e18 + 10;
long long MOD = 1e9 + 7;
const long long MAXN = 2e5 + 5;
void code_idhr_h() {
  long long N;
  cin >> N;
  string A, B;
  cin >> A >> B;
  vector<long long> ansA, ansB;
  for (long long i = 0; i < N - 1; i++) {
    if (A[i] != A[i + 1]) ansA.push_back(i + 1);
  }
  if (A[N - 1] == '1') ansA.push_back(N);
  for (long long i = 0; i < N - 1; i++) {
    if (B[i] != B[i + 1]) ansB.push_back(i + 1);
  }
  if (B[N - 1] == '1') ansB.push_back(N);
  for (auto it = ansB.rbegin(); it != ansB.rend(); it++) ansA.push_back(*it);
  cout << (long long)ansA.size() << " ";
  for (auto it : ansA) cout << it << " ";
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long T = 1;
  cin >> T;
  for (long long t = 1; t <= T; t++) {
    code_idhr_h();
  }
  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;
long long MX = INT_MAX;
long long MN = INT_MIN;
long long INF = 1e18 + 10;
long long MOD = 1e9 + 7;
const long long MAXN = 2e5 + 5;
void code_idhr_h() {
  long long N;
  cin >> N;
  string A, B;
  cin >> A >> B;
  vector<long long> ansA, ansB;
  for (long long i = 0; i < N - 1; i++) {
    if (A[i] != A[i + 1]) ansA.push_back(i + 1);
  }
  if (A[N - 1] == '1') ansA.push_back(N);
  for (long long i = 0; i < N - 1; i++) {
    if (B[i] != B[i + 1]) ansB.push_back(i + 1);
  }
  if (B[N - 1] == '1') ansB.push_back(N);
  for (auto it = ansB.rbegin(); it != ansB.rend(); it++) ansA.push_back(*it);
  cout << (long long)ansA.size() << " ";
  for (auto it : ansA) cout << it << " ";
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long T = 1;
  cin >> T;
  for (long long t = 1; t <= T; t++) {
    code_idhr_h();
  }
  return 0;
}

```