#include <bits/stdc++.h>
using namespace std;
inline long long add(long long a, long long b) {
  return ((a % 1000000007LL) + (b % 1000000007LL)) % 1000000007LL;
}
inline long long mul(long long a, long long b) {
  return ((a % 1000000007LL) * (b % 1000000007LL)) % 1000000007LL;
}
inline long long sub(long long a, long long b) {
  return ((a % 1000000007LL) - (b % 1000000007LL) + 1000000007LL) %
         1000000007LL;
}
void prefixInversionAndSwap(string &ss, long long x) {
  for (long long i = 0; i <= x; i++) {
    if (ss[i] == '0') {
      ss[i] = '1';
    } else {
      ss[i] = '0';
    }
  }
  for (long long i = 0; i <= (x / 2); i++) {
    swap(ss[i], ss[x - i]);
  }
  return;
}
void Flash() {
  long long n;
  cin >> n;
  string a, b;
  cin >> a >> b;
  vector<long long> ans;
  long long swapped = 0;
  long long ind = 0, nai, bbit;
  long long firstBit = a[0] - '0';
  for (long long i = (n - 1); i > 0; i--) {
    if (swapped) {
      nai = (a[ind - i] - '0');
      nai ^= 1;
    } else {
      nai = (a[ind + i] - '0');
    }
    bbit = b[i] - '0';
    if (nai != bbit) {
      if (nai == 1) {
        if (firstBit == 0) {
          ans.push_back(1);
        }
        ans.push_back(i + 1);
        if (swapped) {
          ind -= i;
        } else {
          ind += i;
        }
        swapped ^= 1;
        firstBit = (nai ^ 1);
      } else {
        if (firstBit == 1) {
          ans.push_back(1);
        }
        ans.push_back(i + 1);
        if (swapped) {
          ind -= i;
        } else {
          ind += i;
        }
        swapped ^= 1;
        firstBit = (nai ^ 1);
      }
    }
  }
  if (firstBit != (b[0] - '0')) {
    ans.push_back(1);
  }
  cout << ans.size() << " ";
  for (auto it : ans) {
    cout << it << " ";
  }
  cout << "\n";
}
int main(int argc, char const *argv[]) {
  ios::sync_with_stdio(0), cin.tie(NULL), cout.tie(NULL);
  ;
  long long t = 1;
  cin >> t;
  for (long long i = 1; i <= t; i++) {
    Flash();
  }
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
inline long long add(long long a, long long b) {
  return ((a % 1000000007LL) + (b % 1000000007LL)) % 1000000007LL;
}
inline long long mul(long long a, long long b) {
  return ((a % 1000000007LL) * (b % 1000000007LL)) % 1000000007LL;
}
inline long long sub(long long a, long long b) {
  return ((a % 1000000007LL) - (b % 1000000007LL) + 1000000007LL) %
         1000000007LL;
}
void prefixInversionAndSwap(string &ss, long long x) {
  for (long long i = 0; i <= x; i++) {
    if (ss[i] == '0') {
      ss[i] = '1';
    } else {
      ss[i] = '0';
    }
  }
  for (long long i = 0; i <= (x / 2); i++) {
    swap(ss[i], ss[x - i]);
  }
  return;
}
void Flash() {
  long long n;
  cin >> n;
  string a, b;
  cin >> a >> b;
  vector<long long> ans;
  long long swapped = 0;
  long long ind = 0, nai, bbit;
  long long firstBit = a[0] - '0';
  for (long long i = (n - 1); i > 0; i--) {
    if (swapped) {
      nai = (a[ind - i] - '0');
      nai ^= 1;
    } else {
      nai = (a[ind + i] - '0');
    }
    bbit = b[i] - '0';
    if (nai != bbit) {
      if (nai == 1) {
        if (firstBit == 0) {
          ans.push_back(1);
        }
        ans.push_back(i + 1);
        if (swapped) {
          ind -= i;
        } else {
          ind += i;
        }
        swapped ^= 1;
        firstBit = (nai ^ 1);
      } else {
        if (firstBit == 1) {
          ans.push_back(1);
        }
        ans.push_back(i + 1);
        if (swapped) {
          ind -= i;
        } else {
          ind += i;
        }
        swapped ^= 1;
        firstBit = (nai ^ 1);
      }
    }
  }
  if (firstBit != (b[0] - '0')) {
    ans.push_back(1);
  }
  cout << ans.size() << " ";
  for (auto it : ans) {
    cout << it << " ";
  }
  cout << "\n";
}
int main(int argc, char const *argv[]) {
  ios::sync_with_stdio(0), cin.tie(NULL), cout.tie(NULL);
  ;
  long long t = 1;
  cin >> t;
  for (long long i = 1; i <= t; i++) {
    Flash();
  }
  return 0;
}

```