#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
void solve(int tc) {
  int n;
  cin >> n;
  string a, b;
  cin >> a >> b;
  vector<int> ans;
  int low = 0, high = n - 1;
  int count = 0;
  int trueptr = n - 1;
  while (trueptr > 0) {
    int end, start;
    if (count % 2 == 0) {
      end = low;
      start = high;
    } else {
      start = low;
      end = high;
    }
    char st = a[start];
    if (count % 2 == 1) {
      st = a[start] == '1' ? '0' : '1';
    }
    if (st == b[trueptr]) {
      trueptr--;
      if (count % 2 == 1) {
        low++;
      } else {
        high--;
      }
      continue;
    }
    if (count % 2 == 1) {
      char ch = a[end] == '1' ? '0' : '1';
      if (ch == b[trueptr]) {
        ans.push_back(1);
      }
      ans.push_back(trueptr + 1);
      high--;
      count++;
    } else {
      if (a[end] == b[trueptr]) {
        ans.push_back(1);
      }
      ans.push_back(trueptr + 1);
      low++;
      count++;
    }
    trueptr--;
  }
  int c;
  if (count % 2 == 0) {
    c = low;
  } else {
    c = high;
  }
  char x = a[c];
  if (count % 2 == 1) {
    x = (a[c] == '1') ? '0' : '1';
  }
  if (x != b[0]) {
    ans.push_back(1);
  }
  cout << ans.size() << " ";
  for (int i : ans) {
    cout << i << " ";
  }
  cout << endl;
}
int main() {
  int t;
  cin >> t;
  for (int i = 1; i <= t; i++) {
    solve(i);
  }
  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;
const int mod = 1e9 + 7;
void solve(int tc) {
  int n;
  cin >> n;
  string a, b;
  cin >> a >> b;
  vector<int> ans;
  int low = 0, high = n - 1;
  int count = 0;
  int trueptr = n - 1;
  while (trueptr > 0) {
    int end, start;
    if (count % 2 == 0) {
      end = low;
      start = high;
    } else {
      start = low;
      end = high;
    }
    char st = a[start];
    if (count % 2 == 1) {
      st = a[start] == '1' ? '0' : '1';
    }
    if (st == b[trueptr]) {
      trueptr--;
      if (count % 2 == 1) {
        low++;
      } else {
        high--;
      }
      continue;
    }
    if (count % 2 == 1) {
      char ch = a[end] == '1' ? '0' : '1';
      if (ch == b[trueptr]) {
        ans.push_back(1);
      }
      ans.push_back(trueptr + 1);
      high--;
      count++;
    } else {
      if (a[end] == b[trueptr]) {
        ans.push_back(1);
      }
      ans.push_back(trueptr + 1);
      low++;
      count++;
    }
    trueptr--;
  }
  int c;
  if (count % 2 == 0) {
    c = low;
  } else {
    c = high;
  }
  char x = a[c];
  if (count % 2 == 1) {
    x = (a[c] == '1') ? '0' : '1';
  }
  if (x != b[0]) {
    ans.push_back(1);
  }
  cout << ans.size() << " ";
  for (int i : ans) {
    cout << i << " ";
  }
  cout << endl;
}
int main() {
  int t;
  cin >> t;
  for (int i = 1; i <= t; i++) {
    solve(i);
  }
  return 0;
}

```