#include <bits/stdc++.h>
int main() {
  long long int t, a = 0, b = 0, c = 0, d, i, j, k = 0, x, y, z, n, l, r, s = 0,
                   m = 0;
  scanf("%lld", &t);
  char ar[100005], br[100005];
  for (j = 0; j < t; j++) {
    scanf("%lld", &a);
    int sr[2 * a], t[a];
    scanf("%s %s", ar, br);
    for (i = 1; i < a; i++) {
      if (ar[i] != ar[i - 1]) sr[c++] = i;
    }
    if (ar[a - 1] != br[a - 1]) sr[c++] = a;
    for (i = a - 1; i >= 1; i--) {
      if (br[i] != br[i - 1]) t[k++] = i;
    }
    for (x = 0; x < k; x++) sr[c++] = t[x];
    printf("%lld ", c);
    for (i = 0; i < c; i++) printf("%d ", sr[i]);
    printf("\n");
    c = k = 0;
  }
  return 0;
}

### Prompt

Generate a CPP solution 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>
int main() {
  long long int t, a = 0, b = 0, c = 0, d, i, j, k = 0, x, y, z, n, l, r, s = 0,
                   m = 0;
  scanf("%lld", &t);
  char ar[100005], br[100005];
  for (j = 0; j < t; j++) {
    scanf("%lld", &a);
    int sr[2 * a], t[a];
    scanf("%s %s", ar, br);
    for (i = 1; i < a; i++) {
      if (ar[i] != ar[i - 1]) sr[c++] = i;
    }
    if (ar[a - 1] != br[a - 1]) sr[c++] = a;
    for (i = a - 1; i >= 1; i--) {
      if (br[i] != br[i - 1]) t[k++] = i;
    }
    for (x = 0; x < k; x++) sr[c++] = t[x];
    printf("%lld ", c);
    for (i = 0; i < c; i++) printf("%d ", sr[i]);
    printf("\n");
    c = k = 0;
  }
  return 0;
}

```