#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int T, n;
char a[N], b[N];
vector<int> ans1, ans2;
void solve(char s[], vector<int> &g) {
  for (int i = 1; i <= n - 1; i++) {
    if (s[i] != s[i + 1]) g.push_back(i);
  }
  if (s[n] == '1') g.push_back(n);
}
int main() {
  ios::sync_with_stdio(false);
  cin >> T;
  while (T--) {
    cin >> n >> a + 1 >> b + 1;
    ans1.clear();
    ans2.clear();
    solve(a, ans1);
    solve(b, ans2);
    int sz1 = ans1.size();
    int sz2 = ans2.size();
    printf("%d", sz1 + sz2);
    for (int i = 0; i < sz1; i++) printf(" %d", ans1[i]);
    for (int i = sz2 - 1; i >= 0; i--) printf(" %d", ans2[i]);
    printf("\n");
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 int N = 1e5 + 10;
int T, n;
char a[N], b[N];
vector<int> ans1, ans2;
void solve(char s[], vector<int> &g) {
  for (int i = 1; i <= n - 1; i++) {
    if (s[i] != s[i + 1]) g.push_back(i);
  }
  if (s[n] == '1') g.push_back(n);
}
int main() {
  ios::sync_with_stdio(false);
  cin >> T;
  while (T--) {
    cin >> n >> a + 1 >> b + 1;
    ans1.clear();
    ans2.clear();
    solve(a, ans1);
    solve(b, ans2);
    int sz1 = ans1.size();
    int sz2 = ans2.size();
    printf("%d", sz1 + sz2);
    for (int i = 0; i < sz1; i++) printf(" %d", ans1[i]);
    for (int i = sz2 - 1; i >= 0; i--) printf(" %d", ans2[i]);
    printf("\n");
  }
  return 0;
}

```