#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
mt19937 mt(736);
void gen_ans(string a, const string &b, vector<int> &ans) {
  int l = 0, r = a.size();
  bool rev = false;
  auto atpos = [&a, &l, &r, &rev](int pos) -> char {
    if (rev) return char(a[r - 1 - pos] ^ '0' ^ '1');
    return a[l + pos];
  };
  auto mov = [&ans, &l, &r, &rev]() {
    rev = !rev;
    ans.push_back(r - l);
  };
  auto sw = [&ans, &a, &l, &r, &rev]() {
    a[rev ? r - 1 : l] ^= '0' ^ '1';
    ans.push_back(1);
  };
  auto nxt = [&l, &r, &rev]() {
    if (rev)
      l++;
    else
      r--;
  };
  while (l < r) {
    if (atpos(r - l - 1) != b[r - l - 1]) {
      if (atpos(0) == b[r - l - 1]) sw();
      mov();
    }
    nxt();
  }
}
void solve_test(istream &cin = std::cin, ostream &cout = std::cout) {
  int n;
  cin >> n;
  string a, b;
  cin >> a >> b;
  vector<int> ans;
  ans.reserve(b.size() * 2);
  gen_ans(a, b, ans);
  cout << ans.size();
  for (auto it : ans) cout << ' ' << it;
  cout << endl;
}
void solve(istream &cin = std::cin, ostream &cout = std::cout) {
  int t;
  cin >> t;
  for (int i = 0; i < t; i++) solve_test(cin, cout);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout << fixed;
  solve();
  return 0;
}

### Prompt

Your challenge is to write 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;
using ll = long long;
using ull = unsigned long long;
mt19937 mt(736);
void gen_ans(string a, const string &b, vector<int> &ans) {
  int l = 0, r = a.size();
  bool rev = false;
  auto atpos = [&a, &l, &r, &rev](int pos) -> char {
    if (rev) return char(a[r - 1 - pos] ^ '0' ^ '1');
    return a[l + pos];
  };
  auto mov = [&ans, &l, &r, &rev]() {
    rev = !rev;
    ans.push_back(r - l);
  };
  auto sw = [&ans, &a, &l, &r, &rev]() {
    a[rev ? r - 1 : l] ^= '0' ^ '1';
    ans.push_back(1);
  };
  auto nxt = [&l, &r, &rev]() {
    if (rev)
      l++;
    else
      r--;
  };
  while (l < r) {
    if (atpos(r - l - 1) != b[r - l - 1]) {
      if (atpos(0) == b[r - l - 1]) sw();
      mov();
    }
    nxt();
  }
}
void solve_test(istream &cin = std::cin, ostream &cout = std::cout) {
  int n;
  cin >> n;
  string a, b;
  cin >> a >> b;
  vector<int> ans;
  ans.reserve(b.size() * 2);
  gen_ans(a, b, ans);
  cout << ans.size();
  for (auto it : ans) cout << ' ' << it;
  cout << endl;
}
void solve(istream &cin = std::cin, ostream &cout = std::cout) {
  int t;
  cin >> t;
  for (int i = 0; i < t; i++) solve_test(cin, cout);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout << fixed;
  solve();
  return 0;
}

```