#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long MOD = 1000000007;
const int INF = 0x3f3f3f3f;
int a[200005], b[200005];
int aa, bb;
int ac, bc;
int sh;
vector<int> ans;
void solve(int n) {
  if (n == -1) return;
  if ((a[ac + aa * n] ^ sh) == b[n]) {
    solve(n - 1);
    return;
  }
  if ((a[ac + aa * 0] ^ sh) == b[n]) {
    ans.push_back(1);
    a[ac + aa * 0] ^= 1;
  }
  ans.push_back(n + 1);
  ac = ac + aa * n;
  sh ^= 1;
  aa *= -1;
  solve(n - 1);
}
void go() {
  int n;
  cin >> n;
  aa = 1;
  ac = 0;
  sh = 0;
  for (int i = 0; i < n; ++i) {
    char c;
    cin >> c;
    if (c == '1')
      a[i] = 1;
    else
      a[i] = 0;
  }
  for (int i = 0; i < n; ++i) {
    char c;
    cin >> c;
    if (c == '1')
      b[i] = 1;
    else
      b[i] = 0;
  }
  ans.clear();
  solve(n - 1);
  cout << ans.size() << ' ';
  for (auto& x : ans) cout << x << ' ';
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    go();
  }
}

### 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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long MOD = 1000000007;
const int INF = 0x3f3f3f3f;
int a[200005], b[200005];
int aa, bb;
int ac, bc;
int sh;
vector<int> ans;
void solve(int n) {
  if (n == -1) return;
  if ((a[ac + aa * n] ^ sh) == b[n]) {
    solve(n - 1);
    return;
  }
  if ((a[ac + aa * 0] ^ sh) == b[n]) {
    ans.push_back(1);
    a[ac + aa * 0] ^= 1;
  }
  ans.push_back(n + 1);
  ac = ac + aa * n;
  sh ^= 1;
  aa *= -1;
  solve(n - 1);
}
void go() {
  int n;
  cin >> n;
  aa = 1;
  ac = 0;
  sh = 0;
  for (int i = 0; i < n; ++i) {
    char c;
    cin >> c;
    if (c == '1')
      a[i] = 1;
    else
      a[i] = 0;
  }
  for (int i = 0; i < n; ++i) {
    char c;
    cin >> c;
    if (c == '1')
      b[i] = 1;
    else
      b[i] = 0;
  }
  ans.clear();
  solve(n - 1);
  cout << ans.size() << ' ';
  for (auto& x : ans) cout << x << ' ';
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    go();
  }
}

```