#include <bits/stdc++.h>
using namespace std;
const int INF = (int)INT_MAX;
const long long INF64 = (long long)1e18;
const long double eps = 1e-9;
const long double pi = 3.14159265358979323846;
double PHI = 1.6180339;
template <class T1>
void deb(T1 e) {
  cout << "-->" << e << "\n";
}
template <class T1, class T2>
void deb(T1 e1, T2 e2) {
  cout << "-->" << e1 << " " << e2 << "\n";
}
template <class T1, class T2, class T3>
void deb(T1 e1, T2 e2, T3 e3) {
  cout << "-->" << e1 << " " << e2 << " " << e3 << "\n";
}
template <class T1, class T2, class T3, class T4>
void deb(T1 e1, T2 e2, T3 e3, T4 e4) {
  cout << "-->" << e1 << " " << e2 << " " << e3 << " " << e4 << "\n";
}
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
void read(vector<int>& v) {
  int N = (int)v.size();
  for (int i = 0; i < N; i++) {
    cin >> v[i];
  }
}
void Answer() {
  int n;
  cin >> n;
  string a, b;
  cin >> a >> b;
  if (a == b) {
    cout << "0"
         << "\n";
    return;
  };
  int r = n - 1;
  bool tog = false;
  vector<int> ans;
  int lll = 0, rrr = n - 1;
  while (r >= 0) {
    char A = (tog ? (a[lll] == '1' ? '0' : '1') : a[lll]);
    char B = (tog ? (a[rrr] == '1' ? '0' : '1') : a[rrr]);
    debug() << " ["
            << "lll"
               ": "
            << (lll)
            << "] "
               " ["
            << "rrr"
               ": "
            << (rrr)
            << "] "
               " ["
            << "r"
               ": "
            << (r)
            << "] "
               " ["
            << "A"
               ": "
            << (A)
            << "] "
               " ["
            << "B"
               ": "
            << (B)
            << "] "
               " ["
            << "b[r]"
               ": "
            << (b[r]) << "] ";
    if ((!r ? A : B) == b[r]) {
      --r;
      if (lll < rrr)
        --rrr;
      else
        ++rrr;
    } else if (A == b[r]) {
      ans.push_back(1);
      ans.push_back(r-- + 1);
      tog = !tog;
      swap(lll, rrr);
      if (lll < rrr)
        --rrr;
      else
        ++rrr;
    } else {
      ans.push_back(r-- + 1);
      tog = !tog;
      swap(lll, rrr);
      if (lll < rrr)
        --rrr;
      else
        ++rrr;
    }
  }
  cout << int((ans).size()) << ' ';
  for (auto x : ans) cout << x << ' ';
  cout << "\n";
  ;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  cin >> t;
  while (t--) {
    Answer();
  }
  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>
using namespace std;
const int INF = (int)INT_MAX;
const long long INF64 = (long long)1e18;
const long double eps = 1e-9;
const long double pi = 3.14159265358979323846;
double PHI = 1.6180339;
template <class T1>
void deb(T1 e) {
  cout << "-->" << e << "\n";
}
template <class T1, class T2>
void deb(T1 e1, T2 e2) {
  cout << "-->" << e1 << " " << e2 << "\n";
}
template <class T1, class T2, class T3>
void deb(T1 e1, T2 e2, T3 e3) {
  cout << "-->" << e1 << " " << e2 << " " << e3 << "\n";
}
template <class T1, class T2, class T3, class T4>
void deb(T1 e1, T2 e2, T3 e3, T4 e4) {
  cout << "-->" << e1 << " " << e2 << " " << e3 << " " << e4 << "\n";
}
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
void read(vector<int>& v) {
  int N = (int)v.size();
  for (int i = 0; i < N; i++) {
    cin >> v[i];
  }
}
void Answer() {
  int n;
  cin >> n;
  string a, b;
  cin >> a >> b;
  if (a == b) {
    cout << "0"
         << "\n";
    return;
  };
  int r = n - 1;
  bool tog = false;
  vector<int> ans;
  int lll = 0, rrr = n - 1;
  while (r >= 0) {
    char A = (tog ? (a[lll] == '1' ? '0' : '1') : a[lll]);
    char B = (tog ? (a[rrr] == '1' ? '0' : '1') : a[rrr]);
    debug() << " ["
            << "lll"
               ": "
            << (lll)
            << "] "
               " ["
            << "rrr"
               ": "
            << (rrr)
            << "] "
               " ["
            << "r"
               ": "
            << (r)
            << "] "
               " ["
            << "A"
               ": "
            << (A)
            << "] "
               " ["
            << "B"
               ": "
            << (B)
            << "] "
               " ["
            << "b[r]"
               ": "
            << (b[r]) << "] ";
    if ((!r ? A : B) == b[r]) {
      --r;
      if (lll < rrr)
        --rrr;
      else
        ++rrr;
    } else if (A == b[r]) {
      ans.push_back(1);
      ans.push_back(r-- + 1);
      tog = !tog;
      swap(lll, rrr);
      if (lll < rrr)
        --rrr;
      else
        ++rrr;
    } else {
      ans.push_back(r-- + 1);
      tog = !tog;
      swap(lll, rrr);
      if (lll < rrr)
        --rrr;
      else
        ++rrr;
    }
  }
  cout << int((ans).size()) << ' ';
  for (auto x : ans) cout << x << ' ';
  cout << "\n";
  ;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  cin >> t;
  while (t--) {
    Answer();
  }
  return 0;
}

```