#include <bits/stdc++.h>
using namespace std;
template <class T>
void smin(T &a, T b) {
  a = min(a, b);
}
template <class T>
void smax(T &a, T b) {
  a = max(a, b);
}
template <class T>
T min(vector<T> &a) {
  return *min_element(a.begin(), a.end());
}
template <class T>
T max(vector<T> &a) {
  return *max_element(a.begin(), a.end());
}
template <class T>
void print(vector<T> &a, string sep) {
  for (T el : a) cout << el << sep;
  cout << endl;
}
char nott(char x, int revs) {
  if (revs % 2 == 0) return x;
  if (x == '0')
    return '1';
  else
    return '0';
}
char charr(char x, int revs) {
  if (revs % 2 == 0)
    return x;
  else
    return nott(x, 1);
}
void solve() {
  int n;
  cin >> n;
  string a, b;
  cin >> a >> b;
  if (n == 1) {
    if (a[0] == b[0]) {
      cout << 0 << endl;
      return;
    } else {
      cout << 1 << ' ' << 1 << endl;
      return;
    }
  }
  vector<int> changes;
  int revs = 0;
  int ctr = 1;
  for (int i = n - 1; i > 0; i--) {
    if (a[0] == b[i]) {
      changes.push_back(1);
      changes.push_back(i + 1);
    } else {
      changes.push_back(i + 1);
    }
    revs++;
    if (revs % 2 == 0) {
      a[0] = nott(a[ctr], revs);
      ctr++;
    } else {
      a[0] = nott(a[n - ctr], revs);
    }
  }
  if (a[0] != b[0]) {
    changes.push_back(1);
  }
  cout << changes.size() << ' ';
  print(changes, " ");
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}

### Prompt

In Cpp, your task is to solve 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;
template <class T>
void smin(T &a, T b) {
  a = min(a, b);
}
template <class T>
void smax(T &a, T b) {
  a = max(a, b);
}
template <class T>
T min(vector<T> &a) {
  return *min_element(a.begin(), a.end());
}
template <class T>
T max(vector<T> &a) {
  return *max_element(a.begin(), a.end());
}
template <class T>
void print(vector<T> &a, string sep) {
  for (T el : a) cout << el << sep;
  cout << endl;
}
char nott(char x, int revs) {
  if (revs % 2 == 0) return x;
  if (x == '0')
    return '1';
  else
    return '0';
}
char charr(char x, int revs) {
  if (revs % 2 == 0)
    return x;
  else
    return nott(x, 1);
}
void solve() {
  int n;
  cin >> n;
  string a, b;
  cin >> a >> b;
  if (n == 1) {
    if (a[0] == b[0]) {
      cout << 0 << endl;
      return;
    } else {
      cout << 1 << ' ' << 1 << endl;
      return;
    }
  }
  vector<int> changes;
  int revs = 0;
  int ctr = 1;
  for (int i = n - 1; i > 0; i--) {
    if (a[0] == b[i]) {
      changes.push_back(1);
      changes.push_back(i + 1);
    } else {
      changes.push_back(i + 1);
    }
    revs++;
    if (revs % 2 == 0) {
      a[0] = nott(a[ctr], revs);
      ctr++;
    } else {
      a[0] = nott(a[n - ctr], revs);
    }
  }
  if (a[0] != b[0]) {
    changes.push_back(1);
  }
  cout << changes.size() << ' ';
  print(changes, " ");
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}

```