#include <bits/stdc++.h>
using namespace std;
void DBG() { cerr << "]\n"; }
template <typename T, typename... Args>
void DBG(T first, Args... args) {
  cerr << first;
  if (sizeof...(args)) cerr << ", ";
  DBG(args...);
}
long long int del_x[4] = {-1, 1, 0, 0};
long long int del_y[4] = {0, 0, 1, -1};
long long int XX[8] = {-1, -1, -1, 0, 1, 1, 1, 0};
long long int YY[8] = {-1, 0, 1, 1, 1, 0, -1, -1};
void solve() {
  long long int n;
  cin >> n;
  string a, b;
  cin >> a >> b;
  vector<long long int> ans;
  for (long long int i = 0; i < n; i++) {
    if (a[i] == '0' && a[i + 1] == '1') {
      long long int j = i + 1;
      ans.push_back(i + 1);
      while (j < n && a[j] == '1') {
        j++;
      }
      i = j - 1;
      ans.push_back(j);
    } else if (a[i] == '1') {
      long long int j = i + 1;
      while (j < n && a[j] == '1') {
        j++;
      }
      i = j - 1;
      ans.push_back(j);
    }
  }
  bool update = false;
  for (long long int i = n - 1; i >= 0; i--) {
    if (b[i] == '1' && !update) {
      ans.push_back(i + 1);
      update = true;
    } else if (b[i] == '1' && update) {
      continue;
    } else if (b[i] == '0' && update) {
      ans.push_back(i + 1);
      update = false;
    }
  }
  cout << ans.size() << " ";
  for (auto ele : ans) {
    cout << ele << " ";
  }
  cout << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long int t = 1;
  cin >> t;
  for (long long int tt = 1; tt <= t; tt++) {
    solve();
  }
}

### Prompt

Your challenge is to write 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 DBG() { cerr << "]\n"; }
template <typename T, typename... Args>
void DBG(T first, Args... args) {
  cerr << first;
  if (sizeof...(args)) cerr << ", ";
  DBG(args...);
}
long long int del_x[4] = {-1, 1, 0, 0};
long long int del_y[4] = {0, 0, 1, -1};
long long int XX[8] = {-1, -1, -1, 0, 1, 1, 1, 0};
long long int YY[8] = {-1, 0, 1, 1, 1, 0, -1, -1};
void solve() {
  long long int n;
  cin >> n;
  string a, b;
  cin >> a >> b;
  vector<long long int> ans;
  for (long long int i = 0; i < n; i++) {
    if (a[i] == '0' && a[i + 1] == '1') {
      long long int j = i + 1;
      ans.push_back(i + 1);
      while (j < n && a[j] == '1') {
        j++;
      }
      i = j - 1;
      ans.push_back(j);
    } else if (a[i] == '1') {
      long long int j = i + 1;
      while (j < n && a[j] == '1') {
        j++;
      }
      i = j - 1;
      ans.push_back(j);
    }
  }
  bool update = false;
  for (long long int i = n - 1; i >= 0; i--) {
    if (b[i] == '1' && !update) {
      ans.push_back(i + 1);
      update = true;
    } else if (b[i] == '1' && update) {
      continue;
    } else if (b[i] == '0' && update) {
      ans.push_back(i + 1);
      update = false;
    }
  }
  cout << ans.size() << " ";
  for (auto ele : ans) {
    cout << ele << " ";
  }
  cout << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long int t = 1;
  cin >> t;
  for (long long int tt = 1; tt <= t; tt++) {
    solve();
  }
}

```