#include <bits/stdc++.h>
using namespace std;
int ertek(char c) {
  if (c == '1')
    return 1;
  else
    return 0;
}
void feladat() {
  int n;
  string s;
  string t;
  cin >> n;
  cin >> s >> t;
  vector<int> lepesek;
  int alap = 0;
  int kezd = 0;
  int veg = n - 1;
  int valodi_ertek;
  for (int i = n - 1; i >= 0; i--) {
    valodi_ertek = abs(ertek(s[veg]) - alap);
    if (valodi_ertek == ertek(t[i])) {
      if (kezd <= veg) {
        veg--;
      } else {
        veg++;
      }
      continue;
    }
    if (abs(ertek(s[kezd]) - alap) != 1 - ertek(t[i])) {
      lepesek.push_back(1);
    }
    lepesek.push_back(abs(veg - kezd) + 1);
    alap = 1 - alap;
    if (kezd <= veg) {
      kezd++;
    } else {
      kezd--;
    }
    swap(kezd, veg);
  }
  cout << lepesek.size() << " ";
  for (int ki : lepesek) {
    cout << ki << " ";
  }
  cout << "\n";
}
int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) feladat();
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 ertek(char c) {
  if (c == '1')
    return 1;
  else
    return 0;
}
void feladat() {
  int n;
  string s;
  string t;
  cin >> n;
  cin >> s >> t;
  vector<int> lepesek;
  int alap = 0;
  int kezd = 0;
  int veg = n - 1;
  int valodi_ertek;
  for (int i = n - 1; i >= 0; i--) {
    valodi_ertek = abs(ertek(s[veg]) - alap);
    if (valodi_ertek == ertek(t[i])) {
      if (kezd <= veg) {
        veg--;
      } else {
        veg++;
      }
      continue;
    }
    if (abs(ertek(s[kezd]) - alap) != 1 - ertek(t[i])) {
      lepesek.push_back(1);
    }
    lepesek.push_back(abs(veg - kezd) + 1);
    alap = 1 - alap;
    if (kezd <= veg) {
      kezd++;
    } else {
      kezd--;
    }
    swap(kezd, veg);
  }
  cout << lepesek.size() << " ";
  for (int ki : lepesek) {
    cout << ki << " ";
  }
  cout << "\n";
}
int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) feladat();
}

```