#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:10000000000")
#pragma GCC optimize("O3")
using namespace std;
const int MOD = 1000000007;
const int INF = 1000000007LL;
const long long INF2 = 1LL << 62LL;
const long double EPS = 1e-9;
const int SIZE = 200010;
mt19937 rng(time(0));
uniform_int_distribution<int> uid(-1000000000, 1000000000);
int flag, u[SIZE], d[SIZE], a[SIZE], n;
int rec(int i) {
  if (i == n - 1) return 1;
  if (u[i] == flag) return d[i];
  u[i] = flag;
  if (a[i] == 1) {
    return d[i] = !rec(i + 1);
  }
  return d[i] = 1;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int tt;
  cin >> tt;
  for (int rep_iter = 0; rep_iter < tt; ++rep_iter) {
    int n;
    cin >> n;
    string a, b;
    cin >> a >> b;
    int l = 0, r = n - 1, inv = 0;
    for (auto& i : a) i -= '0';
    for (auto& i : b) i -= '0';
    vector<int> ans;
    for (int len = n; len >= 1; --len) {
      if ((a[r] ^ inv) != b[len - 1]) {
        if ((a[l] ^ inv) == b[len - 1]) {
          ans.push_back(1);
          a[l] = !a[l];
        }
        ans.push_back(len);
        inv = !inv;
        swap(r, l);
      }
      if (r >= l)
        --r;
      else
        ++r;
    }
    cout << ans.size() << ' ';
    for (auto& i : ans) cout << i << ' ';
    cout << '\n';
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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>
#pragma comment(linker, "/STACK:10000000000")
#pragma GCC optimize("O3")
using namespace std;
const int MOD = 1000000007;
const int INF = 1000000007LL;
const long long INF2 = 1LL << 62LL;
const long double EPS = 1e-9;
const int SIZE = 200010;
mt19937 rng(time(0));
uniform_int_distribution<int> uid(-1000000000, 1000000000);
int flag, u[SIZE], d[SIZE], a[SIZE], n;
int rec(int i) {
  if (i == n - 1) return 1;
  if (u[i] == flag) return d[i];
  u[i] = flag;
  if (a[i] == 1) {
    return d[i] = !rec(i + 1);
  }
  return d[i] = 1;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int tt;
  cin >> tt;
  for (int rep_iter = 0; rep_iter < tt; ++rep_iter) {
    int n;
    cin >> n;
    string a, b;
    cin >> a >> b;
    int l = 0, r = n - 1, inv = 0;
    for (auto& i : a) i -= '0';
    for (auto& i : b) i -= '0';
    vector<int> ans;
    for (int len = n; len >= 1; --len) {
      if ((a[r] ^ inv) != b[len - 1]) {
        if ((a[l] ^ inv) == b[len - 1]) {
          ans.push_back(1);
          a[l] = !a[l];
        }
        ans.push_back(len);
        inv = !inv;
        swap(r, l);
      }
      if (r >= l)
        --r;
      else
        ++r;
    }
    cout << ans.size() << ' ';
    for (auto& i : ans) cout << i << ' ';
    cout << '\n';
  }
  return 0;
}

```