#include <bits/stdc++.h>
using namespace std;
const long long inf = 1e9 + 123;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    string s1, s2;
    cin >> n >> s1 >> s2;
    vector<int> ans;
    for (int i = 1; i < n; ++i) {
      if (s1[i] != s1[i - 1]) {
        ans.push_back(i - 1);
      }
    }
    if (s1[n - 1] == '1') ans.push_back(n - 1);
    stack<int> revans;
    for (int i = 1; i < n; ++i) {
      if (s2[i] != s2[i - 1]) {
        revans.push(i - 1);
      }
    }
    if (s2[n - 1] == '1') revans.push(n - 1);
    while (!revans.empty()) {
      ans.push_back(revans.top());
      revans.pop();
    }
    cout << ans.size();
    for (int v : ans) {
      cout << " " << v + 1;
    }
    cout << endl;
  }
}

### Prompt

Your task is to create 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;
const long long inf = 1e9 + 123;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    string s1, s2;
    cin >> n >> s1 >> s2;
    vector<int> ans;
    for (int i = 1; i < n; ++i) {
      if (s1[i] != s1[i - 1]) {
        ans.push_back(i - 1);
      }
    }
    if (s1[n - 1] == '1') ans.push_back(n - 1);
    stack<int> revans;
    for (int i = 1; i < n; ++i) {
      if (s2[i] != s2[i - 1]) {
        revans.push(i - 1);
      }
    }
    if (s2[n - 1] == '1') revans.push(n - 1);
    while (!revans.empty()) {
      ans.push_back(revans.top());
      revans.pop();
    }
    cout << ans.size();
    for (int v : ans) {
      cout << " " << v + 1;
    }
    cout << endl;
  }
}

```