#include <bits/stdc++.h>
using namespace std;
int value(char ch, int flip) {
  int x = ch - '0';
  if (flip == 0)
    return x;
  else
    return 1 - x;
}
void update(int& flip, int& first, int& last) {
  int tmp = last;
  last = first + (last > first ? 1 : -1);
  first = tmp;
  flip = (flip + 1) % 2;
}
vector<int> convert(int n, string a, string b) {
  if (n == 1) {
    if (a[0] != b[0]) {
      return vector<int>{1};
    } else
      return {};
  }
  vector<int> res;
  int first = 0, last = n - 1, flip = 0;
  for (int i = n - 1; i >= 0; --i) {
    if ((i == 0) && value(a[last], flip) != int(b[i] - '0')) {
      res.push_back(1);
      continue;
    }
    if (value(a[last], flip) != int(b[i] - '0')) {
      if (1 - value(a[first], flip) != int(b[i] - '0')) {
        res.push_back(1);
      }
      res.push_back(i + 1);
      update(flip, first, last);
    } else
      last = last + (last > first ? -1 : +1);
  }
  return res;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    string a, b;
    cin >> a;
    cin >> b;
    vector<int> res = convert(n, a, b);
    cout << res.size() << " ";
    for (int i : res) cout << i << " ";
    cout << "\n";
  }
  return 0;
}

### Prompt

Please formulate 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;
int value(char ch, int flip) {
  int x = ch - '0';
  if (flip == 0)
    return x;
  else
    return 1 - x;
}
void update(int& flip, int& first, int& last) {
  int tmp = last;
  last = first + (last > first ? 1 : -1);
  first = tmp;
  flip = (flip + 1) % 2;
}
vector<int> convert(int n, string a, string b) {
  if (n == 1) {
    if (a[0] != b[0]) {
      return vector<int>{1};
    } else
      return {};
  }
  vector<int> res;
  int first = 0, last = n - 1, flip = 0;
  for (int i = n - 1; i >= 0; --i) {
    if ((i == 0) && value(a[last], flip) != int(b[i] - '0')) {
      res.push_back(1);
      continue;
    }
    if (value(a[last], flip) != int(b[i] - '0')) {
      if (1 - value(a[first], flip) != int(b[i] - '0')) {
        res.push_back(1);
      }
      res.push_back(i + 1);
      update(flip, first, last);
    } else
      last = last + (last > first ? -1 : +1);
  }
  return res;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    string a, b;
    cin >> a;
    cin >> b;
    vector<int> res = convert(n, a, b);
    cout << res.size() << " ";
    for (int i : res) cout << i << " ";
    cout << "\n";
  }
  return 0;
}

```