#include <bits/stdc++.h>
using namespace std;
void fun(string a, vector<int> &ans) {
  int cnt1 = 0, cnt0 = 0;
  int n = a.length();
  for (int i = 0; i < n; i++) {
    if (a[i] == '1')
      cnt1++;
    else
      cnt0++;
  }
  if (cnt1 > cnt0) {
    ans.push_back(n);
    string tm = a;
    for (int i = 0; i < n; i++) {
      a[i] = (1 - (tm[n - 1 - i] - '0')) + '0';
    }
  }
  if (a[0] == '1') {
    ans.push_back(1);
    a[0] = '0';
  }
  for (int i = 1; i < n; i++) {
    if (a[i] == '1') {
      ans.push_back(i);
      ans.push_back(i + 1);
    }
  }
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    string a, b;
    cin >> a >> b;
    vector<int> ans;
    fun(a, ans);
    vector<int> ans2;
    fun(b, ans2);
    reverse(ans2.begin(), ans2.end());
    for (int i = 0; i < ans2.size(); i++) {
      ans.push_back(ans2[i]);
    }
    cout << ans.size() << " ";
    for (int i = 0; i < ans.size(); i++) {
      cout << ans[i] << " ";
    }
    cout << endl;
  }
}

### 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;
void fun(string a, vector<int> &ans) {
  int cnt1 = 0, cnt0 = 0;
  int n = a.length();
  for (int i = 0; i < n; i++) {
    if (a[i] == '1')
      cnt1++;
    else
      cnt0++;
  }
  if (cnt1 > cnt0) {
    ans.push_back(n);
    string tm = a;
    for (int i = 0; i < n; i++) {
      a[i] = (1 - (tm[n - 1 - i] - '0')) + '0';
    }
  }
  if (a[0] == '1') {
    ans.push_back(1);
    a[0] = '0';
  }
  for (int i = 1; i < n; i++) {
    if (a[i] == '1') {
      ans.push_back(i);
      ans.push_back(i + 1);
    }
  }
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    string a, b;
    cin >> a >> b;
    vector<int> ans;
    fun(a, ans);
    vector<int> ans2;
    fun(b, ans2);
    reverse(ans2.begin(), ans2.end());
    for (int i = 0; i < ans2.size(); i++) {
      ans.push_back(ans2[i]);
    }
    cout << ans.size() << " ";
    for (int i = 0; i < ans.size(); i++) {
      cout << ans[i] << " ";
    }
    cout << endl;
  }
}

```