#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
using namespace std;
bool per_square(long double a) {
  if (a < 0) return false;
  long long sr = sqrt(a);
  return (sr * sr == a);
}
bool is_prime(long long x) {
  for (long long i = 2; i < sqrt(x) + 1; i++)
    if (x % i == 0) return false;
  return true;
}
bool sort_by_sec(const pair<long long, long long> &a,
                 const pair<long long, long long> &b) {
  return (a.second < b.second);
}
template <typename T, typename T1>
T amax(T &a, T1 b) {
  if (b > a) a = b;
  return a;
}
template <typename T, typename T1>
T amin(T &a, T1 b) {
  if (b < a) a = b;
  return a;
}
void solve() {
  long long n;
  cin >> n;
  string a, b;
  cin >> a >> b;
  vector<long long> ops1, ops2;
  a += '0';
  b += '0';
  for (long long i = 1; i < n + 1; i++) {
    if (a[i] != a[i - 1]) ops1.push_back(i);
    if (b[i] != b[i - 1]) ops2.push_back(i);
  }
  ops1.insert(ops1.end(), ops2.rbegin(), ops2.rend());
  cout << ops1.size() << " ";
  for (auto &i : ops1) cout << i << " ";
  cout << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long tests = 1;
  cin >> tests;
  for (long long test = 1; test <= tests; test++) {
    solve();
  }
  return 0;
}

### Prompt

Develop a solution in cpp 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("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
using namespace std;
bool per_square(long double a) {
  if (a < 0) return false;
  long long sr = sqrt(a);
  return (sr * sr == a);
}
bool is_prime(long long x) {
  for (long long i = 2; i < sqrt(x) + 1; i++)
    if (x % i == 0) return false;
  return true;
}
bool sort_by_sec(const pair<long long, long long> &a,
                 const pair<long long, long long> &b) {
  return (a.second < b.second);
}
template <typename T, typename T1>
T amax(T &a, T1 b) {
  if (b > a) a = b;
  return a;
}
template <typename T, typename T1>
T amin(T &a, T1 b) {
  if (b < a) a = b;
  return a;
}
void solve() {
  long long n;
  cin >> n;
  string a, b;
  cin >> a >> b;
  vector<long long> ops1, ops2;
  a += '0';
  b += '0';
  for (long long i = 1; i < n + 1; i++) {
    if (a[i] != a[i - 1]) ops1.push_back(i);
    if (b[i] != b[i - 1]) ops2.push_back(i);
  }
  ops1.insert(ops1.end(), ops2.rbegin(), ops2.rend());
  cout << ops1.size() << " ";
  for (auto &i : ops1) cout << i << " ";
  cout << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long tests = 1;
  cin >> tests;
  for (long long test = 1; test <= tests; test++) {
    solve();
  }
  return 0;
}

```