#include <bits/stdc++.h>
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n;
    scanf("%d", &n);
    int i, x1, x2, ct1 = 0, ct2 = 0, arr[n];
    int *ch1, *ch2;
    ch1 = (int *)malloc(sizeof(int));
    ch2 = (int *)malloc(sizeof(int));
    scanf("%1d", &arr[0]);
    x1 = arr[0];
    for (i = 1; i < n; i++) {
      scanf("%1d", &arr[i]);
      if (arr[i] != arr[i - 1]) {
        ct1++;
        ch1 = (int *)realloc(ch1, ct1 * sizeof(int));
        ch1[ct1 - 1] = i;
        x1 = arr[i];
      }
    }
    int arr2[n];
    scanf("%1d", &arr2[0]);
    x2 = arr2[0];
    for (i = 1; i < n; i++) {
      scanf("%1d", &arr2[i]);
      if (arr2[i] != arr2[i - 1]) {
        ct2++;
        ch2 = (int *)realloc(ch2, ct2 * sizeof(int));
        ch2[ct2 - 1] = i;
        x2 = arr2[i];
      }
    }
    if (x1 == arr2[n - 1])
      printf("%d ", ct1 + ct2);
    else
      printf("%d ", ct1 + ct2 + 1);
    for (i = 0; i < ct1; i++) printf("%d ", ch1[i]);
    if (x1 != arr2[n - 1]) printf("%d ", n);
    for (i = ct2 - 1; i >= 0; i--) printf("%d ", ch2[i]);
    printf("\n");
  }
}

### Prompt

In cpp, your task is to solve 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>
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n;
    scanf("%d", &n);
    int i, x1, x2, ct1 = 0, ct2 = 0, arr[n];
    int *ch1, *ch2;
    ch1 = (int *)malloc(sizeof(int));
    ch2 = (int *)malloc(sizeof(int));
    scanf("%1d", &arr[0]);
    x1 = arr[0];
    for (i = 1; i < n; i++) {
      scanf("%1d", &arr[i]);
      if (arr[i] != arr[i - 1]) {
        ct1++;
        ch1 = (int *)realloc(ch1, ct1 * sizeof(int));
        ch1[ct1 - 1] = i;
        x1 = arr[i];
      }
    }
    int arr2[n];
    scanf("%1d", &arr2[0]);
    x2 = arr2[0];
    for (i = 1; i < n; i++) {
      scanf("%1d", &arr2[i]);
      if (arr2[i] != arr2[i - 1]) {
        ct2++;
        ch2 = (int *)realloc(ch2, ct2 * sizeof(int));
        ch2[ct2 - 1] = i;
        x2 = arr2[i];
      }
    }
    if (x1 == arr2[n - 1])
      printf("%d ", ct1 + ct2);
    else
      printf("%d ", ct1 + ct2 + 1);
    for (i = 0; i < ct1; i++) printf("%d ", ch1[i]);
    if (x1 != arr2[n - 1]) printf("%d ", n);
    for (i = ct2 - 1; i >= 0; i--) printf("%d ", ch2[i]);
    printf("\n");
  }
}

```