#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int inf = 1061109567;
const ll INF = 4557430888798830399;
const int MOD = 1e9 + 7;
int n;
vector<int> calc(string s) {
  s += "0";
  s = " " + s;
  vector<int> res;
  for (int i = 1; i <= n; ++i) {
    if (s[i] != s[i + 1]) res.emplace_back(i);
  }
  return res;
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  if (fopen(""
            ".in",
            "r")) {
    freopen(
        ""
        ".in",
        "r", stdin);
    freopen(
        ""
        ".out",
        "w", stdout);
  }
  int tc;
  cin >> tc;
  while (tc--) {
    string a, b;
    cin >> n >> a >> b;
    vector<int> res = calc(a);
    vector<int> ans = calc(b);
    cout << (int)res.size() + (int)ans.size() << " ";
    for (auto v : res) cout << v << " ";
    reverse(ans.begin(), ans.end());
    for (auto v : ans) cout << v << " ";
    cout << "\n";
  }
  return 0;
}

### Prompt

Please formulate 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;
using ll = long long;
const int inf = 1061109567;
const ll INF = 4557430888798830399;
const int MOD = 1e9 + 7;
int n;
vector<int> calc(string s) {
  s += "0";
  s = " " + s;
  vector<int> res;
  for (int i = 1; i <= n; ++i) {
    if (s[i] != s[i + 1]) res.emplace_back(i);
  }
  return res;
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  if (fopen(""
            ".in",
            "r")) {
    freopen(
        ""
        ".in",
        "r", stdin);
    freopen(
        ""
        ".out",
        "w", stdout);
  }
  int tc;
  cin >> tc;
  while (tc--) {
    string a, b;
    cin >> n >> a >> b;
    vector<int> res = calc(a);
    vector<int> ans = calc(b);
    cout << (int)res.size() + (int)ans.size() << " ";
    for (auto v : res) cout << v << " ";
    reverse(ans.begin(), ans.end());
    for (auto v : ans) cout << v << " ";
    cout << "\n";
  }
  return 0;
}

```