#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void gi(T &x) {
  char ch;
  x = 0;
  int f = 1;
  while (isspace(ch = getchar()))
    ;
  if (ch == '-') ch = getchar(), f = -1;
  do x = x * 10 + ch - '0';
  while (isdigit(ch = getchar()));
  x *= f;
}
template <class T>
void upmax(T &x, T y) {
  x = x > y ? x : y;
}
template <class T>
void upmin(T &x, T y) {
  x = x < y ? x : y;
}
const int N = 100005;
char a[N], b[N];
int main() {
  int T;
  gi(T);
  while (T--) {
    int n;
    gi(n);
    int flip = 0, bd[2];
    bd[0] = 1;
    bd[1] = n;
    scanf("%s%s", a + 1, b + 1);
    for (int i = 1; i <= n; i++) a[i] -= '0', b[i] -= '0';
    static int op[N * 2];
    int oc = 0;
    for (int i = n; i > 1; i--) {
      int x = a[bd[flip ^ 1]] ^ flip, y = b[i];
      if (x == y) {
        if (flip ^ 1)
          --bd[flip ^ 1];
        else
          ++bd[flip ^ 1];
        continue;
      }
      if ((a[bd[flip]] ^ flip) == y) op[++oc] = 1;
      op[++oc] = i;
      flip ^= 1;
      if (flip ^ 1)
        --bd[flip ^ 1];
      else
        ++bd[flip ^ 1];
    }
    if ((a[bd[1]] ^ flip) != b[1]) op[++oc] = 1;
    printf("%d ", oc);
    for (int i = 1; i <= oc; i++) printf("%d ", op[i]);
    puts("");
  }
  return 0;
}

### Prompt

Your challenge is to write 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>
using namespace std;
template <typename T>
inline void gi(T &x) {
  char ch;
  x = 0;
  int f = 1;
  while (isspace(ch = getchar()))
    ;
  if (ch == '-') ch = getchar(), f = -1;
  do x = x * 10 + ch - '0';
  while (isdigit(ch = getchar()));
  x *= f;
}
template <class T>
void upmax(T &x, T y) {
  x = x > y ? x : y;
}
template <class T>
void upmin(T &x, T y) {
  x = x < y ? x : y;
}
const int N = 100005;
char a[N], b[N];
int main() {
  int T;
  gi(T);
  while (T--) {
    int n;
    gi(n);
    int flip = 0, bd[2];
    bd[0] = 1;
    bd[1] = n;
    scanf("%s%s", a + 1, b + 1);
    for (int i = 1; i <= n; i++) a[i] -= '0', b[i] -= '0';
    static int op[N * 2];
    int oc = 0;
    for (int i = n; i > 1; i--) {
      int x = a[bd[flip ^ 1]] ^ flip, y = b[i];
      if (x == y) {
        if (flip ^ 1)
          --bd[flip ^ 1];
        else
          ++bd[flip ^ 1];
        continue;
      }
      if ((a[bd[flip]] ^ flip) == y) op[++oc] = 1;
      op[++oc] = i;
      flip ^= 1;
      if (flip ^ 1)
        --bd[flip ^ 1];
      else
        ++bd[flip ^ 1];
    }
    if ((a[bd[1]] ^ flip) != b[1]) op[++oc] = 1;
    printf("%d ", oc);
    for (int i = 1; i <= oc; i++) printf("%d ", op[i]);
    puts("");
  }
  return 0;
}

```