#include <bits/stdc++.h>
using namespace std;
namespace zzc {
const int maxn = 1e5 + 5;
int t, n, ans1[maxn], ans2[maxn], cnt1, cnt2;
char a[maxn], b[maxn];
void work() {
  scanf("%d", &t);
  while (t--) {
    cnt1 = 0;
    cnt2 = 0;
    scanf("%d", &n);
    scanf("%s", a + 1);
    scanf("%s", b + 1);
    for (int i = 1; i < n; i++) {
      if (a[i] != a[i + 1]) {
        ans1[++cnt1] = i;
      }
    }
    for (int i = 1; i < n; i++) {
      if (b[i] != b[i + 1]) {
        ans2[++cnt2] = i;
      }
    }
    if (a[n] == b[n])
      printf("%d ", cnt1 + cnt2);
    else
      printf("%d ", cnt1 + cnt2 + 1);
    for (int i = 1; i <= cnt1; i++) printf("%d ", ans1[i]);
    if (a[n] != b[n]) printf("%d ", n);
    for (int i = cnt2; i >= 1; i--) printf("%d ", ans2[i]);
    puts(" ");
  }
}
}  // namespace zzc
int main() {
  zzc::work();
  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;
namespace zzc {
const int maxn = 1e5 + 5;
int t, n, ans1[maxn], ans2[maxn], cnt1, cnt2;
char a[maxn], b[maxn];
void work() {
  scanf("%d", &t);
  while (t--) {
    cnt1 = 0;
    cnt2 = 0;
    scanf("%d", &n);
    scanf("%s", a + 1);
    scanf("%s", b + 1);
    for (int i = 1; i < n; i++) {
      if (a[i] != a[i + 1]) {
        ans1[++cnt1] = i;
      }
    }
    for (int i = 1; i < n; i++) {
      if (b[i] != b[i + 1]) {
        ans2[++cnt2] = i;
      }
    }
    if (a[n] == b[n])
      printf("%d ", cnt1 + cnt2);
    else
      printf("%d ", cnt1 + cnt2 + 1);
    for (int i = 1; i <= cnt1; i++) printf("%d ", ans1[i]);
    if (a[n] != b[n]) printf("%d ", n);
    for (int i = cnt2; i >= 1; i--) printf("%d ", ans2[i]);
    puts(" ");
  }
}
}  // namespace zzc
int main() {
  zzc::work();
  return 0;
}

```