#include <bits/stdc++.h>
using namespace std;
unsigned long long int powmod(long long int a, long long int b) {
  if (b == 0) return 1;
  if (b % 2 == 0) {
    unsigned long long int p = powmod(a, b / 2) % 1000000007;
    return (p * p) % 1000000007;
  }
  return (a * powmod(a, b - 1)) % 1000000007;
}
long long int expmod(long long int a, long long int b) {
  long long int rlt = 1;
  for (a %= 1000000007; b; a = a * a % 1000000007, b >>= 1)
    if (b & 1) rlt = rlt * a % 1000000007;
  return rlt;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long int lcm(long long int a, long long int b) {
  long long int g = gcd(a, b);
  return (a * b) / g;
}
long long int sieve_of_eratosthenes(long long int n) {
  bool check[n];
  long long int i;
  memset(check, 0, sizeof(check));
  for (i = 2; i <= n; i++) {
    if (check[i] == 0) {
      for (long long int j = i * 2; j <= n; j += i) check[j] = 1;
    }
  }
  i = n;
  while (check[i] != 0) i--;
  return i;
}
inline int power(int a, int b) {
  int x = 1;
  while (b) {
    if (b & 1) x *= a;
    a *= a;
    b >>= 1;
  }
  return x;
}
long long int max(long long int a, long long int b) { return a > b ? a : b; }
long long int min(long long int a, long long int b) { return a < b ? a : b; }
void solve() {
  int n;
  cin >> n;
  string a, b;
  cin >> a >> b;
  if (a == b) {
    cout << 0 << "\n";
    return;
  }
  vector<int> v;
  for (int i = 0; i < n; i++) {
    if (a[i] == '1') {
      if (i != 0) v.push_back(i);
      while (i < n && a[i] == '1') i++;
      v.push_back(i);
      i--;
    }
  }
  int count = 0;
  for (auto i : b) count += (i == '1');
  if (count > 0) {
    string c = b;
    reverse(b.begin(), b.end());
    bool apply = 0;
    for (int i = n - 1; i >= 0; --i) {
      int c = 0;
      int start = i;
      while (i >= 0 && b[i] == '0') {
        c++;
        i--;
      }
      if (start == i) continue;
      v.push_back(start + 1);
      if (i >= 0) v.push_back(i + 1);
      i++;
    }
    v.push_back(n);
  }
  vector<int> moves;
  for (int i = 0; i < v.size(); i++) {
    if (moves.size() && v[i] == moves[moves.size() - 1])
      moves.pop_back();
    else
      moves.push_back(v[i]);
  }
  cout << moves.size() << " ";
  for (auto i : moves) cout << i << " ";
  cout << "\n";
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) solve();
  return 0;
}

### 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>
using namespace std;
unsigned long long int powmod(long long int a, long long int b) {
  if (b == 0) return 1;
  if (b % 2 == 0) {
    unsigned long long int p = powmod(a, b / 2) % 1000000007;
    return (p * p) % 1000000007;
  }
  return (a * powmod(a, b - 1)) % 1000000007;
}
long long int expmod(long long int a, long long int b) {
  long long int rlt = 1;
  for (a %= 1000000007; b; a = a * a % 1000000007, b >>= 1)
    if (b & 1) rlt = rlt * a % 1000000007;
  return rlt;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long int lcm(long long int a, long long int b) {
  long long int g = gcd(a, b);
  return (a * b) / g;
}
long long int sieve_of_eratosthenes(long long int n) {
  bool check[n];
  long long int i;
  memset(check, 0, sizeof(check));
  for (i = 2; i <= n; i++) {
    if (check[i] == 0) {
      for (long long int j = i * 2; j <= n; j += i) check[j] = 1;
    }
  }
  i = n;
  while (check[i] != 0) i--;
  return i;
}
inline int power(int a, int b) {
  int x = 1;
  while (b) {
    if (b & 1) x *= a;
    a *= a;
    b >>= 1;
  }
  return x;
}
long long int max(long long int a, long long int b) { return a > b ? a : b; }
long long int min(long long int a, long long int b) { return a < b ? a : b; }
void solve() {
  int n;
  cin >> n;
  string a, b;
  cin >> a >> b;
  if (a == b) {
    cout << 0 << "\n";
    return;
  }
  vector<int> v;
  for (int i = 0; i < n; i++) {
    if (a[i] == '1') {
      if (i != 0) v.push_back(i);
      while (i < n && a[i] == '1') i++;
      v.push_back(i);
      i--;
    }
  }
  int count = 0;
  for (auto i : b) count += (i == '1');
  if (count > 0) {
    string c = b;
    reverse(b.begin(), b.end());
    bool apply = 0;
    for (int i = n - 1; i >= 0; --i) {
      int c = 0;
      int start = i;
      while (i >= 0 && b[i] == '0') {
        c++;
        i--;
      }
      if (start == i) continue;
      v.push_back(start + 1);
      if (i >= 0) v.push_back(i + 1);
      i++;
    }
    v.push_back(n);
  }
  vector<int> moves;
  for (int i = 0; i < v.size(); i++) {
    if (moves.size() && v[i] == moves[moves.size() - 1])
      moves.pop_back();
    else
      moves.push_back(v[i]);
  }
  cout << moves.size() << " ";
  for (auto i : moves) cout << i << " ";
  cout << "\n";
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) solve();
  return 0;
}

```