#include <bits/stdc++.h>
using namespace std;
int t, n, epos, spos, j, ans;
int main() {
  cin >> t;
  while (t--) {
    cin >> n;
    string a, b;
    cin >> a;
    cin >> b;
    bool state = false;
    epos = n - 1;
    spos = 0;
    j = n - 1;
    vector<int> ans;
    while (j >= 0) {
      bool flag = false;
      if (state) {
        flag = (b[j] - '0') == !(a[epos] - '0');
      } else
        flag = b[j] == a[epos];
      if (!flag) {
        if (state) {
          if ((b[j] - '0') == !(a[spos] - '0')) ans.push_back(1);
          ans.push_back(abs(epos - spos) + 1);
        } else {
          if (b[j] == a[spos]) ans.push_back(1);
          ans.push_back(abs(epos - spos) + 1);
        }
        state = !state;
        if (spos < epos) {
          int tmp = epos;
          epos = spos + 1;
          spos = tmp;
        } else {
          int tmp = epos;
          epos = spos - 1;
          spos = tmp;
        }
      } else {
        if (spos > epos)
          epos++;
        else
          epos--;
      }
      j--;
    }
    cout << ans.size() << ' ';
    for (auto i : ans) cout << i << ' ';
    cout << endl;
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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 t, n, epos, spos, j, ans;
int main() {
  cin >> t;
  while (t--) {
    cin >> n;
    string a, b;
    cin >> a;
    cin >> b;
    bool state = false;
    epos = n - 1;
    spos = 0;
    j = n - 1;
    vector<int> ans;
    while (j >= 0) {
      bool flag = false;
      if (state) {
        flag = (b[j] - '0') == !(a[epos] - '0');
      } else
        flag = b[j] == a[epos];
      if (!flag) {
        if (state) {
          if ((b[j] - '0') == !(a[spos] - '0')) ans.push_back(1);
          ans.push_back(abs(epos - spos) + 1);
        } else {
          if (b[j] == a[spos]) ans.push_back(1);
          ans.push_back(abs(epos - spos) + 1);
        }
        state = !state;
        if (spos < epos) {
          int tmp = epos;
          epos = spos + 1;
          spos = tmp;
        } else {
          int tmp = epos;
          epos = spos - 1;
          spos = tmp;
        }
      } else {
        if (spos > epos)
          epos++;
        else
          epos--;
      }
      j--;
    }
    cout << ans.size() << ' ';
    for (auto i : ans) cout << i << ' ';
    cout << endl;
  }
  return 0;
}

```