#include <bits/stdc++.h>
using namespace std;
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long tt;
  cin >> tt;
  while (tt--) {
    long long n;
    cin >> n;
    string a, b;
    cin >> a >> b;
    long long l = n - 1;
    long long k = 0;
    vector<long long> res;
    long long c = 0;
    long long ll = 0, rr = n - 1;
    for (long long i = n - 1; i >= 0; i--) {
      long long x;
      if ((n - 1 - i) % 2 == 0)
        x = ll;
      else
        x = rr;
      if (c % 2 == 0) {
        if (a[x] != b[i]) {
          res.push_back(i + 1);
          c++;
        } else {
          k++;
          res.push_back(1);
          c++;
          res.push_back(i + 1);
        }
      } else {
        if (a[x] == b[i]) {
          res.push_back(i + 1);
          c++;
        } else {
          k++;
          res.push_back(1);
          c++;
          res.push_back(i + 1);
        }
      }
      if ((n - 1 - i) % 2 == 0)
        ll++;
      else
        rr--;
    }
    cout << k + c << " ";
    for (long long i = 0; i < res.size(); i++) cout << res[i] << " ";
    cout << endl;
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long tt;
  cin >> tt;
  while (tt--) {
    long long n;
    cin >> n;
    string a, b;
    cin >> a >> b;
    long long l = n - 1;
    long long k = 0;
    vector<long long> res;
    long long c = 0;
    long long ll = 0, rr = n - 1;
    for (long long i = n - 1; i >= 0; i--) {
      long long x;
      if ((n - 1 - i) % 2 == 0)
        x = ll;
      else
        x = rr;
      if (c % 2 == 0) {
        if (a[x] != b[i]) {
          res.push_back(i + 1);
          c++;
        } else {
          k++;
          res.push_back(1);
          c++;
          res.push_back(i + 1);
        }
      } else {
        if (a[x] == b[i]) {
          res.push_back(i + 1);
          c++;
        } else {
          k++;
          res.push_back(1);
          c++;
          res.push_back(i + 1);
        }
      }
      if ((n - 1 - i) % 2 == 0)
        ll++;
      else
        rr--;
    }
    cout << k + c << " ";
    for (long long i = 0; i < res.size(); i++) cout << res[i] << " ";
    cout << endl;
  }
  return 0;
}

```