#include <bits/stdc++.h>
using namespace std;
void swap(string &a, int i, vector<int> &A) {
  A.push_back(i + 1);
  for (int j = 0; j <= i / 2; j++) {
    if (a[j] == a[i - j]) {
      if (a[j] == '0') {
        a[j] = '1';
        a[i - j] = '1';
      } else {
        a[j] = '0';
        a[i - j] = '0';
      }
    }
  }
  return;
}
void solve(string a, vector<int> &A) {
  int i = 0;
  char p = a[0];
  int n = a.length();
  while (i < n - 1) {
    while (i < n && a[i] == p) {
      i++;
    }
    if (i < n) {
      A.push_back(i);
      if (p == '0') {
        p = '1';
      } else {
        p = '0';
      }
    }
  }
  if (p == '0') {
    A.push_back(n);
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  for (int t = 0; t < T; t++) {
    int n;
    string a, b;
    cin >> n >> a >> b;
    vector<int> A;
    vector<int> B;
    solve(a, A);
    solve(b, B);
    int m = A.size();
    int k = B.size();
    cout << m + k << " ";
    for (int i = 0; i < m; i++) {
      cout << A[i] << " ";
    }
    for (int i = k - 1; i >= 0; i--) {
      cout << B[i] << " ";
    }
    cout << endl;
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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 swap(string &a, int i, vector<int> &A) {
  A.push_back(i + 1);
  for (int j = 0; j <= i / 2; j++) {
    if (a[j] == a[i - j]) {
      if (a[j] == '0') {
        a[j] = '1';
        a[i - j] = '1';
      } else {
        a[j] = '0';
        a[i - j] = '0';
      }
    }
  }
  return;
}
void solve(string a, vector<int> &A) {
  int i = 0;
  char p = a[0];
  int n = a.length();
  while (i < n - 1) {
    while (i < n && a[i] == p) {
      i++;
    }
    if (i < n) {
      A.push_back(i);
      if (p == '0') {
        p = '1';
      } else {
        p = '0';
      }
    }
  }
  if (p == '0') {
    A.push_back(n);
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  for (int t = 0; t < T; t++) {
    int n;
    string a, b;
    cin >> n >> a >> b;
    vector<int> A;
    vector<int> B;
    solve(a, A);
    solve(b, B);
    int m = A.size();
    int k = B.size();
    cout << m + k << " ";
    for (int i = 0; i < m; i++) {
      cout << A[i] << " ";
    }
    for (int i = k - 1; i >= 0; i--) {
      cout << B[i] << " ";
    }
    cout << endl;
  }
  return 0;
}

```