#include <bits/stdc++.h>
using namespace std;
priority_queue<long long int, vector<long long int>, greater<long long int>> ti;
vector<long long int> p[200005], b(200005, 0), d(200005, 0), f;
map<pair<long long int, long long int>, long long int> mp;
set<pair<long long int, long long int>> s;
map<long long int, long long int> np, v;
queue<long long int> q;
vector<vector<long long int>> a(3, vector<long long int>(100005, 0));
long long int ans = 0;
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int bpow(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % 1000000007;
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return res % 1000000007;
}
long long int fact(long long int i) {
  if (i <= 0)
    return 1;
  else
    return (i * fact(i - 1)) % 1000000007;
}
long long int isprime(long long int n) {
  for (long long int i = 2; i <= sqrt(n); i++)
    if (n % i == 0) return 0;
  return 1;
}
bool cmp(long long int i, long long int j) { return p[i].size() < p[j].size(); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    string a, b;
    cin >> a >> b;
    vector<long long int> p, q;
    for (long long int i = 0; i < n - 1; i++) {
      if (a[i] != a[i + 1]) p.push_back(i);
    }
    if (a[n - 1] == '1') p.push_back(n - 1);
    for (long long int i = 0; i < n - 1; i++) {
      if (b[i] != b[i + 1]) q.push_back(i);
    }
    if (b[n - 1] == '1') q.push_back(n - 1);
    cout << p.size() + q.size() << "\n";
    for (long long int i = 0; i < p.size(); i++) cout << p[i] + 1 << " ";
    for (long long int i = q.size() - 1; i >= 0; i--) {
      cout << q[i] + 1 << " ";
    }
    cout << "\n";
  }
}

### Prompt

Your challenge is to write 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;
priority_queue<long long int, vector<long long int>, greater<long long int>> ti;
vector<long long int> p[200005], b(200005, 0), d(200005, 0), f;
map<pair<long long int, long long int>, long long int> mp;
set<pair<long long int, long long int>> s;
map<long long int, long long int> np, v;
queue<long long int> q;
vector<vector<long long int>> a(3, vector<long long int>(100005, 0));
long long int ans = 0;
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int bpow(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % 1000000007;
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return res % 1000000007;
}
long long int fact(long long int i) {
  if (i <= 0)
    return 1;
  else
    return (i * fact(i - 1)) % 1000000007;
}
long long int isprime(long long int n) {
  for (long long int i = 2; i <= sqrt(n); i++)
    if (n % i == 0) return 0;
  return 1;
}
bool cmp(long long int i, long long int j) { return p[i].size() < p[j].size(); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    string a, b;
    cin >> a >> b;
    vector<long long int> p, q;
    for (long long int i = 0; i < n - 1; i++) {
      if (a[i] != a[i + 1]) p.push_back(i);
    }
    if (a[n - 1] == '1') p.push_back(n - 1);
    for (long long int i = 0; i < n - 1; i++) {
      if (b[i] != b[i + 1]) q.push_back(i);
    }
    if (b[n - 1] == '1') q.push_back(n - 1);
    cout << p.size() + q.size() << "\n";
    for (long long int i = 0; i < p.size(); i++) cout << p[i] + 1 << " ";
    for (long long int i = q.size() - 1; i >= 0; i--) {
      cout << q[i] + 1 << " ";
    }
    cout << "\n";
  }
}

```