#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
using namespace std;
long long int mod = 1000000007;
long long int maxn = 200001;
long long int binpow(long long int a, long long int b, long long int mod) {
  long long int res = 1;
  while (b > 0) {
    if (b % 2) {
      res = (res * a) % mod;
      b--;
    }
    a = (a * a) % mod;
    b /= 2;
  }
  return res % mod;
}
long long int binpow(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b % 2) {
      res = (res * a);
      b--;
    }
    a = (a * a);
    b /= 2;
  }
  return res;
}
bool isPrime(long long int n) {
  if (n == 1) return false;
  for (long long int i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
long long int mulmod(long long int a, long long int b) {
  return ((a % mod) * (b % mod)) % mod;
}
long long int addmod(long long int a, long long int b) {
  return (((a % mod) + (b % mod)) + mod) % mod;
}
void solve();
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int t = 2;
  cin >> t;
  for (long long int tt = 1; tt <= t; tt++) {
    solve();
  }
}
void solve() {
  long long int n;
  cin >> n;
  string s1, s2;
  cin >> s1 >> s2;
  vector<long long int> oper;
  bool zero;
  if (s1[0] == '0')
    zero = true;
  else
    zero = false;
  for (long long int i = 0; i < n - 1; i++) {
    if (s1[i] != s1[i + 1]) {
      oper.push_back(i + 1);
      zero = !zero;
    }
  }
  if (!zero) oper.push_back(n);
  s1 = string(n, '0');
  zero = true;
  for (long long int i = n - 1; i >= 0; i--) {
    if (s2[i] == '1' and zero == true) {
      oper.push_back(i + 1);
      zero = false;
    }
    if (s2[i] == '0' and zero == false) {
      oper.push_back(i + 1);
      zero = true;
    }
  }
  cout << oper.size() << " ";
  for (long long int i = 0; i < oper.size(); i++) cout << oper[i] << " ";
  cout << "\n";
}

### 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>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
using namespace std;
long long int mod = 1000000007;
long long int maxn = 200001;
long long int binpow(long long int a, long long int b, long long int mod) {
  long long int res = 1;
  while (b > 0) {
    if (b % 2) {
      res = (res * a) % mod;
      b--;
    }
    a = (a * a) % mod;
    b /= 2;
  }
  return res % mod;
}
long long int binpow(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b % 2) {
      res = (res * a);
      b--;
    }
    a = (a * a);
    b /= 2;
  }
  return res;
}
bool isPrime(long long int n) {
  if (n == 1) return false;
  for (long long int i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
long long int mulmod(long long int a, long long int b) {
  return ((a % mod) * (b % mod)) % mod;
}
long long int addmod(long long int a, long long int b) {
  return (((a % mod) + (b % mod)) + mod) % mod;
}
void solve();
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int t = 2;
  cin >> t;
  for (long long int tt = 1; tt <= t; tt++) {
    solve();
  }
}
void solve() {
  long long int n;
  cin >> n;
  string s1, s2;
  cin >> s1 >> s2;
  vector<long long int> oper;
  bool zero;
  if (s1[0] == '0')
    zero = true;
  else
    zero = false;
  for (long long int i = 0; i < n - 1; i++) {
    if (s1[i] != s1[i + 1]) {
      oper.push_back(i + 1);
      zero = !zero;
    }
  }
  if (!zero) oper.push_back(n);
  s1 = string(n, '0');
  zero = true;
  for (long long int i = n - 1; i >= 0; i--) {
    if (s2[i] == '1' and zero == true) {
      oper.push_back(i + 1);
      zero = false;
    }
    if (s2[i] == '0' and zero == false) {
      oper.push_back(i + 1);
      zero = true;
    }
  }
  cout << oper.size() << " ";
  for (long long int i = 0; i < oper.size(); i++) cout << oper[i] << " ";
  cout << "\n";
}

```