#include <bits/stdc++.h>
using namespace std;
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
bool cmp(pair<long long, long long> &a, pair<long long, long long> &b) {
  if (a.second == b.second) return a.first < b.first;
  return a.second < b.second;
}
const long long MOD = 1e9 + 7;
long long powerLL(long long x, long long n) {
  long long result = 1;
  while (n) {
    if (n & 1) result = result * x % MOD;
    n = n / 2;
    x = x * x % MOD;
  }
  return result;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n, m, i, j, k, f = 0, sum = 0, ans = 0, x, y, z;
    string s, s1;
    cin >> n >> s >> s1;
    vector<long long> v;
    for (i = 1; i <= n - 1; i++) {
      if (s[i] != s[i - 1]) v.push_back(i);
    }
    char ch = s[n - 1];
    for (i = n - 1; i >= 0; i--) {
      if (ch != s1[i]) {
        v.push_back(i + 1);
        if (ch == '1')
          ch = '0';
        else
          ch = '1';
      }
    }
    cout << v.size() << " ";
    for (auto it : v) cout << it << " ";
    cout << "\n";
  }
  return 0;
}

### Prompt

Please formulate 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;
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
bool cmp(pair<long long, long long> &a, pair<long long, long long> &b) {
  if (a.second == b.second) return a.first < b.first;
  return a.second < b.second;
}
const long long MOD = 1e9 + 7;
long long powerLL(long long x, long long n) {
  long long result = 1;
  while (n) {
    if (n & 1) result = result * x % MOD;
    n = n / 2;
    x = x * x % MOD;
  }
  return result;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n, m, i, j, k, f = 0, sum = 0, ans = 0, x, y, z;
    string s, s1;
    cin >> n >> s >> s1;
    vector<long long> v;
    for (i = 1; i <= n - 1; i++) {
      if (s[i] != s[i - 1]) v.push_back(i);
    }
    char ch = s[n - 1];
    for (i = n - 1; i >= 0; i--) {
      if (ch != s1[i]) {
        v.push_back(i + 1);
        if (ch == '1')
          ch = '0';
        else
          ch = '1';
      }
    }
    cout << v.size() << " ";
    for (auto it : v) cout << it << " ";
    cout << "\n";
  }
  return 0;
}

```