#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long int t, n, c[(int)2e5 + 2];
  char a[(int)2e5 + 2], b[(int)2e5 + 2];
  cin >> t;
  while (t--) {
    long int k = 0, x = 0, y = 0, e = 0, j = 0;
    cin >> n >> a >> b;
    for (int i = 0; i < n; i++) {
      if (a[i] == '1' && e == 1 || a[i] == '1' && i == 0) {
        x++;
        e = 1;
        if (i == n - 1) {
          c[j] = n;
          j++;
        }
      } else if (a[i] == '0' && e == 0 || a[i] == '0' && i == 0) {
        y++;
        e = 0;
      } else if (a[i] == '0' && e == 1) {
        c[j] = i;
        j++;
        x = 0;
        y++;
        e = 0;
      } else if (a[i] == '1' && e == 0) {
        c[j] = i;
        j++;
        y = 0;
        x++;
        e = 1;
        if (i == n - 1) {
          c[j] = n;
          j++;
        }
      }
    }
    x = 0;
    y = 0;
    e = 0;
    int p = 0;
    if (b[n - 1] == '1') {
      c[j] = n;
      j++;
    }
    for (int i = n - 1; i >= 0; i--) {
      if (b[i] == '1' && e == 1 || b[i] == '1' && i == n - 1) {
        x++;
        e = 1;
      } else if (b[i] == '0' && e == 0 || b[i] == '0' && i == n - 1) {
        y++;
        e = 0;
      } else if (b[i] == '0' && e == 1) {
        c[j] = i + 1;
        j++;
        x = 0;
        y++;
        e = 0;
      } else if (b[i] == '1' && e == 0) {
        c[j] = i + 1;
        j++;
        y = 0;
        x++;
        e = 1;
      }
    }
    cout << j << " ";
    for (int i = 0; i < j; i++) cout << c[i] << " ";
    cout << endl;
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long int t, n, c[(int)2e5 + 2];
  char a[(int)2e5 + 2], b[(int)2e5 + 2];
  cin >> t;
  while (t--) {
    long int k = 0, x = 0, y = 0, e = 0, j = 0;
    cin >> n >> a >> b;
    for (int i = 0; i < n; i++) {
      if (a[i] == '1' && e == 1 || a[i] == '1' && i == 0) {
        x++;
        e = 1;
        if (i == n - 1) {
          c[j] = n;
          j++;
        }
      } else if (a[i] == '0' && e == 0 || a[i] == '0' && i == 0) {
        y++;
        e = 0;
      } else if (a[i] == '0' && e == 1) {
        c[j] = i;
        j++;
        x = 0;
        y++;
        e = 0;
      } else if (a[i] == '1' && e == 0) {
        c[j] = i;
        j++;
        y = 0;
        x++;
        e = 1;
        if (i == n - 1) {
          c[j] = n;
          j++;
        }
      }
    }
    x = 0;
    y = 0;
    e = 0;
    int p = 0;
    if (b[n - 1] == '1') {
      c[j] = n;
      j++;
    }
    for (int i = n - 1; i >= 0; i--) {
      if (b[i] == '1' && e == 1 || b[i] == '1' && i == n - 1) {
        x++;
        e = 1;
      } else if (b[i] == '0' && e == 0 || b[i] == '0' && i == n - 1) {
        y++;
        e = 0;
      } else if (b[i] == '0' && e == 1) {
        c[j] = i + 1;
        j++;
        x = 0;
        y++;
        e = 0;
      } else if (b[i] == '1' && e == 0) {
        c[j] = i + 1;
        j++;
        y = 0;
        x++;
        e = 1;
      }
    }
    cout << j << " ";
    for (int i = 0; i < j; i++) cout << c[i] << " ";
    cout << endl;
  }
  return 0;
}

```