#include <bits/stdc++.h>
const long long P = 1e9 + 7;
const double EPS = 1e-9;
using namespace std;
void flip(string &a, int x) {
  for (int i = 0; i < x; i++) {
    if (a[i] == '1')
      a[i] = '0';
    else
      a[i] = '1';
  }
  reverse(a.begin(), a.begin() + x);
}
bool validate(string &a, string &b, vector<int> &res) {
  for (auto x : res) {
    flip(a, x);
  }
  return a == b;
}
void solve() {
  int n;
  cin >> n;
  string a, b;
  cin >> a >> b;
  vector<int> res;
  if (a[0] != b[0]) {
    res.push_back(1);
  }
  int i = 1;
  while (i < n) {
    if (a[i] == b[i]) {
      i++;
      continue;
    }
    if (i + 1 < n) {
      int f = a[i], s = a[i + 1];
      int bac = i + 2;
      if (s == b[i + 1]) {
        res.push_back(i + 1);
        res.push_back(1);
        res.push_back(i + 1);
      } else if (f == b[i + 1] && s == b[i]) {
        res.push_back(bac - 1);
        res.push_back(bac);
        res.push_back(bac - 1);
        res.push_back(bac - 2);
      } else {
        res.push_back(bac);
        res.push_back(2);
        res.push_back(bac);
      }
      i += 2;
    } else {
      res.push_back(i + 1);
      res.push_back(1);
      res.push_back(i + 1);
      i++;
    }
  }
  cout << res.size() << " ";
  if (res.size() > 0) {
    for (auto x : res) cout << x << " ";
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}

### 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>
const long long P = 1e9 + 7;
const double EPS = 1e-9;
using namespace std;
void flip(string &a, int x) {
  for (int i = 0; i < x; i++) {
    if (a[i] == '1')
      a[i] = '0';
    else
      a[i] = '1';
  }
  reverse(a.begin(), a.begin() + x);
}
bool validate(string &a, string &b, vector<int> &res) {
  for (auto x : res) {
    flip(a, x);
  }
  return a == b;
}
void solve() {
  int n;
  cin >> n;
  string a, b;
  cin >> a >> b;
  vector<int> res;
  if (a[0] != b[0]) {
    res.push_back(1);
  }
  int i = 1;
  while (i < n) {
    if (a[i] == b[i]) {
      i++;
      continue;
    }
    if (i + 1 < n) {
      int f = a[i], s = a[i + 1];
      int bac = i + 2;
      if (s == b[i + 1]) {
        res.push_back(i + 1);
        res.push_back(1);
        res.push_back(i + 1);
      } else if (f == b[i + 1] && s == b[i]) {
        res.push_back(bac - 1);
        res.push_back(bac);
        res.push_back(bac - 1);
        res.push_back(bac - 2);
      } else {
        res.push_back(bac);
        res.push_back(2);
        res.push_back(bac);
      }
      i += 2;
    } else {
      res.push_back(i + 1);
      res.push_back(1);
      res.push_back(i + 1);
      i++;
    }
  }
  cout << res.size() << " ";
  if (res.size() > 0) {
    for (auto x : res) cout << x << " ";
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}

```