#include <bits/stdc++.h>
using namespace std;
using ll = long long;
void scan(int& a) { scanf("%d", &a); }
void scan(long long& a) { scanf("%lld", &a); }
template <class T>
void scan(T& a) {
  cin >> a;
}
template <class T>
void scan(vector<T>& vec) {
  for (auto&& it : vec) scan(it);
}
void in() {}
template <class Head, class... Tail>
void in(Head& head, Tail&... tail) {
  scan(head);
  in(tail...);
}
void print(const int& a) { printf("%d", a); }
void print(const long long& a) { printf("%lld", a); }
void print(const double& a) { printf("%.15lf", a); }
template <class T>
void print(const T& a) {
  cout << a;
}
template <class T>
void print(const vector<T>& vec) {
  if (vec.empty()) return;
  print(vec[0]);
  for (auto it = vec.begin(); ++it != vec.end();) {
    putchar(' ');
    print(*it);
  }
}
void out() { putchar('\n'); }
template <class T>
void out(const T& t) {
  print(t);
  putchar('\n');
}
template <class Head, class... Tail>
void out(const Head& head, const Tail&... tail) {
  print(head);
  putchar(' ');
  out(tail...);
}
template <class T>
void dprint(const T& a) {
  cerr << a;
}
template <class T>
void dprint(const vector<T>& vec) {
  if (vec.empty()) return;
  cerr << vec[0];
  for (auto it = vec.begin(); ++it != vec.end();) {
    cerr << " " << *it;
  }
}
void debug() { cerr << endl; }
template <class T>
void debug(const T& t) {
  dprint(t);
  cerr << endl;
}
template <class Head, class... Tail>
void debug(const Head& head, const Tail&... tail) {
  dprint(head);
  cerr << " ";
  debug(tail...);
}
int main() {
  int n;
  string a, b;
  int test;
  scanf("%d", &test);
  while (test--) {
    in(n, a, b);
    vector<int> ope;
    int f = 0, r = n - 1, rev = 0;
    for (int i = n - 1; i >= 0; i--) {
      if (rev % 2 == 0) {
        if (a[r] != b[i]) {
          if (a[f] == b[i]) {
            ope.push_back(1);
            if (a[f] == '1')
              a[f] = '0';
            else
              a[f] = '1';
          }
          if (a[f] != b[i]) {
            ope.push_back(i + 1);
            rev++;
            f++;
          }
        } else {
          r--;
        }
      } else {
        if (a[f] == b[i]) {
          if (a[r] != b[i]) {
            ope.push_back(1);
            if (a[r] == '1')
              a[r] = '0';
            else
              a[r] = '1';
          }
          if (a[r] == b[i]) {
            ope.push_back(i + 1);
            rev++;
            r--;
          }
        } else {
          f++;
        }
      }
    }
    out(ope.size(), ope);
  }
}

### Prompt

Your task is to create 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;
using ll = long long;
void scan(int& a) { scanf("%d", &a); }
void scan(long long& a) { scanf("%lld", &a); }
template <class T>
void scan(T& a) {
  cin >> a;
}
template <class T>
void scan(vector<T>& vec) {
  for (auto&& it : vec) scan(it);
}
void in() {}
template <class Head, class... Tail>
void in(Head& head, Tail&... tail) {
  scan(head);
  in(tail...);
}
void print(const int& a) { printf("%d", a); }
void print(const long long& a) { printf("%lld", a); }
void print(const double& a) { printf("%.15lf", a); }
template <class T>
void print(const T& a) {
  cout << a;
}
template <class T>
void print(const vector<T>& vec) {
  if (vec.empty()) return;
  print(vec[0]);
  for (auto it = vec.begin(); ++it != vec.end();) {
    putchar(' ');
    print(*it);
  }
}
void out() { putchar('\n'); }
template <class T>
void out(const T& t) {
  print(t);
  putchar('\n');
}
template <class Head, class... Tail>
void out(const Head& head, const Tail&... tail) {
  print(head);
  putchar(' ');
  out(tail...);
}
template <class T>
void dprint(const T& a) {
  cerr << a;
}
template <class T>
void dprint(const vector<T>& vec) {
  if (vec.empty()) return;
  cerr << vec[0];
  for (auto it = vec.begin(); ++it != vec.end();) {
    cerr << " " << *it;
  }
}
void debug() { cerr << endl; }
template <class T>
void debug(const T& t) {
  dprint(t);
  cerr << endl;
}
template <class Head, class... Tail>
void debug(const Head& head, const Tail&... tail) {
  dprint(head);
  cerr << " ";
  debug(tail...);
}
int main() {
  int n;
  string a, b;
  int test;
  scanf("%d", &test);
  while (test--) {
    in(n, a, b);
    vector<int> ope;
    int f = 0, r = n - 1, rev = 0;
    for (int i = n - 1; i >= 0; i--) {
      if (rev % 2 == 0) {
        if (a[r] != b[i]) {
          if (a[f] == b[i]) {
            ope.push_back(1);
            if (a[f] == '1')
              a[f] = '0';
            else
              a[f] = '1';
          }
          if (a[f] != b[i]) {
            ope.push_back(i + 1);
            rev++;
            f++;
          }
        } else {
          r--;
        }
      } else {
        if (a[f] == b[i]) {
          if (a[r] != b[i]) {
            ope.push_back(1);
            if (a[r] == '1')
              a[r] = '0';
            else
              a[r] = '1';
          }
          if (a[r] == b[i]) {
            ope.push_back(i + 1);
            rev++;
            r--;
          }
        } else {
          f++;
        }
      }
    }
    out(ope.size(), ope);
  }
}

```