#include <bits/stdc++.h>
using namespace std;
long long poww(long long a, long long b, long long md) {
  return (!b ? 1
             : (b & 1 ? a * poww(a * a % md, b / 2, md) % md
                      : poww(a * a % md, b / 2, md) % md));
}
const long long MAXN = 1e6 + 10;
const long long INF = 8e18;
const long long MOD = 1e9 + 7;
long long B[MAXN], n, s = 0;
long long D(long long x) { return (x + s) % 2; }
long long solve() {
  cin >> n;
  s = 0;
  deque<long long> A;
  for (int i = 0; i < n; i++) {
    char c;
    cin >> c;
    if (c == '1')
      A.push_back(1);
    else
      A.push_back(0);
  }
  for (int i = 0; i < n; i++) {
    char c;
    cin >> c;
    if (c == '1')
      B[i] = 1;
    else
      B[i] = 0;
  }
  vector<long long> ans;
  bool bc = true;
  for (int i = n - 1; i >= 0; i--) {
    if (bc) {
      if (D(A.back()) != B[i]) {
        if (D(A.front()) == B[i]) ans.push_back(1);
        ans.push_back(i + 1);
        A.pop_front();
        bc = !bc;
        s++;
      } else
        A.pop_back();
    } else {
      if (D(A.front()) != B[i]) {
        if (D(A.back()) == B[i]) ans.push_back(1);
        ans.push_back(i + 1);
        A.pop_back();
        bc = !bc;
        s++;
      } else
        A.pop_front();
    }
  }
  cout << ans.size() << ' ';
  for (long long e : ans) cout << e << ' ';
  cout << '\n';
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t;
  cin >> t;
  while (t--) solve();
  return 0;
}

### Prompt

Please formulate 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;
long long poww(long long a, long long b, long long md) {
  return (!b ? 1
             : (b & 1 ? a * poww(a * a % md, b / 2, md) % md
                      : poww(a * a % md, b / 2, md) % md));
}
const long long MAXN = 1e6 + 10;
const long long INF = 8e18;
const long long MOD = 1e9 + 7;
long long B[MAXN], n, s = 0;
long long D(long long x) { return (x + s) % 2; }
long long solve() {
  cin >> n;
  s = 0;
  deque<long long> A;
  for (int i = 0; i < n; i++) {
    char c;
    cin >> c;
    if (c == '1')
      A.push_back(1);
    else
      A.push_back(0);
  }
  for (int i = 0; i < n; i++) {
    char c;
    cin >> c;
    if (c == '1')
      B[i] = 1;
    else
      B[i] = 0;
  }
  vector<long long> ans;
  bool bc = true;
  for (int i = n - 1; i >= 0; i--) {
    if (bc) {
      if (D(A.back()) != B[i]) {
        if (D(A.front()) == B[i]) ans.push_back(1);
        ans.push_back(i + 1);
        A.pop_front();
        bc = !bc;
        s++;
      } else
        A.pop_back();
    } else {
      if (D(A.front()) != B[i]) {
        if (D(A.back()) == B[i]) ans.push_back(1);
        ans.push_back(i + 1);
        A.pop_back();
        bc = !bc;
        s++;
      } else
        A.pop_front();
    }
  }
  cout << ans.size() << ' ';
  for (long long e : ans) cout << e << ' ';
  cout << '\n';
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t;
  cin >> t;
  while (t--) solve();
  return 0;
}

```