#include <bits/stdc++.h>
namespace IO {
template <typename T>
inline void read(T &x) {
  x = 0;
  int f = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
  x *= f;
  return;
}
template <typename T>
inline void write(T x = 0) {
  T wr[51];
  wr[0] = 0;
  if (x < 0) putchar('-'), x = -x;
  if (!x) putchar(48);
  while (x) wr[++wr[0]] = x % 10, x /= 10;
  while (wr[0]) putchar(48 + wr[wr[0]--]);
  return;
}
}  // namespace IO
using IO::read;
using IO::write;
using namespace std;
const int N = 1e6 + 11;
int T, n, now;
int a[N], b[N], ans[N];
int main() {
  read(T);
  while (T--) {
    ans[0] = 0;
    read(n);
    for (register int i = 1; i <= n; ++i) scanf("%01d", a + i);
    for (register int i = 1; i <= n; ++i) scanf("%01d", b + i);
    a[n + 1] = 0;
    b[n + 1] = 0;
    for (register int i = 1; i <= n; ++i)
      if (a[i] != a[i + 1]) ans[++ans[0]] = i;
    for (register int i = n; i >= 1; --i)
      if (b[i] != b[i + 1]) ans[++ans[0]] = i;
    for (register int i = 0; i <= ans[0]; ++i) printf("%d ", ans[i]);
    puts("");
  }
  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>
namespace IO {
template <typename T>
inline void read(T &x) {
  x = 0;
  int f = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
  x *= f;
  return;
}
template <typename T>
inline void write(T x = 0) {
  T wr[51];
  wr[0] = 0;
  if (x < 0) putchar('-'), x = -x;
  if (!x) putchar(48);
  while (x) wr[++wr[0]] = x % 10, x /= 10;
  while (wr[0]) putchar(48 + wr[wr[0]--]);
  return;
}
}  // namespace IO
using IO::read;
using IO::write;
using namespace std;
const int N = 1e6 + 11;
int T, n, now;
int a[N], b[N], ans[N];
int main() {
  read(T);
  while (T--) {
    ans[0] = 0;
    read(n);
    for (register int i = 1; i <= n; ++i) scanf("%01d", a + i);
    for (register int i = 1; i <= n; ++i) scanf("%01d", b + i);
    a[n + 1] = 0;
    b[n + 1] = 0;
    for (register int i = 1; i <= n; ++i)
      if (a[i] != a[i + 1]) ans[++ans[0]] = i;
    for (register int i = n; i >= 1; --i)
      if (b[i] != b[i + 1]) ans[++ans[0]] = i;
    for (register int i = 0; i <= ans[0]; ++i) printf("%d ", ans[i]);
    puts("");
  }
  return 0;
}

```