#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
const int N = 2e5 + 100;
const int maxn = 1e6;
const int MAXL = sqrt(1e9) + 1;
inline int read() {
  char c = getchar();
  int x = 0, f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  return x * f;
}
int n;
char s[N], t[N];
int a[2 * N], len = 0;
void solve() {
  n = read();
  len = 0;
  scanf("%s%s", s + 1, t + 1);
  for (int i = 2; i <= n; i++)
    if (s[i] != s[i - 1]) a[++len] = i - 1;
  int c = s[n] - '0';
  for (int i = n; i >= 1; i--) {
    if (t[i] - '0' != c) {
      a[++len] = i;
      c ^= 1;
    }
  }
  printf("%d ", len);
  for (int i = 1; i <= len; i++) printf("%d ", a[i]);
  printf("\n");
}
int main() {
  int T;
  scanf("%d", &T);
  int tt = 0;
  while (T--) {
    solve();
  }
  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>
using namespace std;
const long long mod = 998244353;
const int N = 2e5 + 100;
const int maxn = 1e6;
const int MAXL = sqrt(1e9) + 1;
inline int read() {
  char c = getchar();
  int x = 0, f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  return x * f;
}
int n;
char s[N], t[N];
int a[2 * N], len = 0;
void solve() {
  n = read();
  len = 0;
  scanf("%s%s", s + 1, t + 1);
  for (int i = 2; i <= n; i++)
    if (s[i] != s[i - 1]) a[++len] = i - 1;
  int c = s[n] - '0';
  for (int i = n; i >= 1; i--) {
    if (t[i] - '0' != c) {
      a[++len] = i;
      c ^= 1;
    }
  }
  printf("%d ", len);
  for (int i = 1; i <= len; i++) printf("%d ", a[i]);
  printf("\n");
}
int main() {
  int T;
  scanf("%d", &T);
  int tt = 0;
  while (T--) {
    solve();
  }
  return 0;
}

```