#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    string a, b;
    cin >> a >> b;
    vector<long long int> cnt;
    int c = 1;
    for (int i = 1; i < n; i += 1) {
      if (a[i] == a[i - 1])
        c++;
      else {
        cnt.emplace_back(c);
        c = 1;
      }
    }
    cnt.emplace_back(c);
    if (a.back() == '0') cnt.pop_back();
    for (auto it = cnt.begin(); it != cnt.end(); ++it) {
      if (it == cnt.begin()) continue;
      *it += *(it - 1);
    }
    vector<long long int> cnt2;
    c = 1;
    for (int i = 1; i < n; i += 1) {
      if (b[i] == b[i - 1])
        c++;
      else {
        cnt2.emplace_back(c);
        c = 1;
      }
    }
    cnt2.emplace_back(c);
    if (b.back() == '0') cnt2.pop_back();
    for (auto it = cnt2.begin(); it != cnt2.end(); ++it) {
      if (it == cnt2.begin()) continue;
      *it += *(it - 1);
    }
    for (auto it = cnt2.rbegin(); it != cnt2.rend(); ++it)
      cnt.emplace_back(*it);
    cout << cnt.size() << ' ';
    for (auto it = cnt.begin(); it != cnt.end(); ++it) cout << *it << ' ';
    cout << '\n';
  }
}

### Prompt

Your task is to create 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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    string a, b;
    cin >> a >> b;
    vector<long long int> cnt;
    int c = 1;
    for (int i = 1; i < n; i += 1) {
      if (a[i] == a[i - 1])
        c++;
      else {
        cnt.emplace_back(c);
        c = 1;
      }
    }
    cnt.emplace_back(c);
    if (a.back() == '0') cnt.pop_back();
    for (auto it = cnt.begin(); it != cnt.end(); ++it) {
      if (it == cnt.begin()) continue;
      *it += *(it - 1);
    }
    vector<long long int> cnt2;
    c = 1;
    for (int i = 1; i < n; i += 1) {
      if (b[i] == b[i - 1])
        c++;
      else {
        cnt2.emplace_back(c);
        c = 1;
      }
    }
    cnt2.emplace_back(c);
    if (b.back() == '0') cnt2.pop_back();
    for (auto it = cnt2.begin(); it != cnt2.end(); ++it) {
      if (it == cnt2.begin()) continue;
      *it += *(it - 1);
    }
    for (auto it = cnt2.rbegin(); it != cnt2.rend(); ++it)
      cnt.emplace_back(*it);
    cout << cnt.size() << ' ';
    for (auto it = cnt.begin(); it != cnt.end(); ++it) cout << *it << ' ';
    cout << '\n';
  }
}

```