#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9 + 7;
const int MOD = 1e9 + 7;
const int MAXN = 1e6 + 3;
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long power(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = res * x;
    y = y >> 1;
    x = x * x;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    string a, b;
    cin >> a >> b;
    long long idx = 0;
    long long turn = 0;
    vector<long long> ans;
    for (long long i = (n - 1); i >= 0; i--) {
      if (turn == 1 && a[idx] != b[i]) {
        ans.push_back(1ll);
      }
      if (turn == 0 && a[idx] == b[i]) {
        ans.push_back(1ll);
      }
      ans.push_back(i + 1);
      if (turn) {
        idx -= i;
      } else
        idx += i;
      turn ^= 1;
    }
    cout << ans.size() << " ";
    for (auto it : ans) cout << it << " ";
    cout << '\n';
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution 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;
const int INF = 1e9 + 7;
const int MOD = 1e9 + 7;
const int MAXN = 1e6 + 3;
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long power(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = res * x;
    y = y >> 1;
    x = x * x;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    string a, b;
    cin >> a >> b;
    long long idx = 0;
    long long turn = 0;
    vector<long long> ans;
    for (long long i = (n - 1); i >= 0; i--) {
      if (turn == 1 && a[idx] != b[i]) {
        ans.push_back(1ll);
      }
      if (turn == 0 && a[idx] == b[i]) {
        ans.push_back(1ll);
      }
      ans.push_back(i + 1);
      if (turn) {
        idx -= i;
      } else
        idx += i;
      turn ^= 1;
    }
    cout << ans.size() << " ";
    for (auto it : ans) cout << it << " ";
    cout << '\n';
  }
  return 0;
}

```