#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
long long int gcd(long long int a, long long int b) {
  return b == 0 ? a : gcd(b, a % b);
}
long long int power(long long int x, long long int n) {
  long long int result = 1;
  while (n > 0) {
    if (n % 2 == 1) result = (result * x) % 1000000007;
    x = (x * x) % 1000000007;
    n = n / 2;
  }
  return result % 1000000007;
}
bool isPrime(long long int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
long long int inv(long long int a, long long int m) { return power(a, m - 2); }
long long int lcm(long long int a, long long int b) {
  return a * b / (gcd(a, b));
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    string str, ch;
    cin >> str;
    cin >> ch;
    bool x = true;
    for (long long int i = 0; i < n; i++) {
      if (str[i] != ch[i]) x = false;
    }
    if (x)
      cout << 0 << '\n';
    else if (n == 1) {
      if (str[0] != ch[0])
        cout << 1 << " " << 1 << '\n';
      else
        cout << 0 << '\n';
    } else {
      vector<long long int> vec;
      long long int x = str[0] - '0';
      for (long long int i = 0; i < n; i++) {
        if ((long long int)(str[i] - '0') == x)
          continue;
        else {
          vec.emplace_back(i);
          x = 1 - x;
        }
      }
      vector<long long int> v;
      for (long long int i = n - 1; i >= 0; i--) {
        if (x == (long long int)(ch[i] - '0'))
          continue;
        else {
          v.emplace_back(i + 1);
          x = 1 - x;
        }
      }
      cout << (long long int)vec.size() + (long long int)v.size() << " ";
      for (long long int i = 0; i < vec.size(); i++) cout << vec[i] << " ";
      for (long long int i = 0; i < (long long int)v.size(); i++)
        cout << v[i] << " ";
      cout << '\n';
    }
  }
}

### 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>
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
long long int gcd(long long int a, long long int b) {
  return b == 0 ? a : gcd(b, a % b);
}
long long int power(long long int x, long long int n) {
  long long int result = 1;
  while (n > 0) {
    if (n % 2 == 1) result = (result * x) % 1000000007;
    x = (x * x) % 1000000007;
    n = n / 2;
  }
  return result % 1000000007;
}
bool isPrime(long long int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
long long int inv(long long int a, long long int m) { return power(a, m - 2); }
long long int lcm(long long int a, long long int b) {
  return a * b / (gcd(a, b));
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    string str, ch;
    cin >> str;
    cin >> ch;
    bool x = true;
    for (long long int i = 0; i < n; i++) {
      if (str[i] != ch[i]) x = false;
    }
    if (x)
      cout << 0 << '\n';
    else if (n == 1) {
      if (str[0] != ch[0])
        cout << 1 << " " << 1 << '\n';
      else
        cout << 0 << '\n';
    } else {
      vector<long long int> vec;
      long long int x = str[0] - '0';
      for (long long int i = 0; i < n; i++) {
        if ((long long int)(str[i] - '0') == x)
          continue;
        else {
          vec.emplace_back(i);
          x = 1 - x;
        }
      }
      vector<long long int> v;
      for (long long int i = n - 1; i >= 0; i--) {
        if (x == (long long int)(ch[i] - '0'))
          continue;
        else {
          v.emplace_back(i + 1);
          x = 1 - x;
        }
      }
      cout << (long long int)vec.size() + (long long int)v.size() << " ";
      for (long long int i = 0; i < vec.size(); i++) cout << vec[i] << " ";
      for (long long int i = 0; i < (long long int)v.size(); i++)
        cout << v[i] << " ";
      cout << '\n';
    }
  }
}

```