#include <bits/stdc++.h>
long long int inf = 1e18;
using namespace std;
long long int add(long long a, long long b) {
  return ((a % 1000000007) + (b % 1000000007)) % 1000000007;
}
long long int subtract(long long a, long long b) {
  return ((a % 1000000007) - (b % 1000000007)) % 1000000007;
}
long long int mult(long long a, long long b) {
  return ((a % 1000000007) * (b % 1000000007)) % 1000000007;
}
long long int add1(long long a, long long b) {
  return ((a % 998244353) + (b % 998244353)) % 998244353;
}
long long int subtract1(long long a, long long b) {
  return ((a % 998244353) - (b % 998244353)) % 998244353;
}
long long int mult1(long long a, long long b) {
  return ((a % 998244353) * (b % 998244353)) % 998244353;
}
long long int binmodwithMOD1(long long int a, long long int b) {
  long long int res = 1;
  while (b) {
    if (b & 1) res = mult1(res, a) % 998244353;
    a = mult1(a, a) % 998244353;
    b >>= 1;
  }
  return res;
}
long long int binmodwithMOD(long long int a, long long int b) {
  long long int res = 1;
  while (b) {
    if (b & 1) res = mult(res, a) % 1000000007;
    a = mult(a, a) % 1000000007;
    b >>= 1;
  }
  return res;
}
long long int binmodwithoutMOD1orMOD2(long long int a, long long int b) {
  long long int res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int inverse(long long int n) {
  return binmodwithMOD(n, 1000000007 - 2);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    string s1, s2;
    cin >> s1 >> s2;
    vector<long long int> a, b;
    for (long long int i = 1; i < n; i++) {
      if (s1[i] != s1[i - 1]) {
        a.push_back(i);
      }
    }
    if (s1[n - 1] != s2[n - 1]) a.push_back(n);
    for (long long int i = 1; i < n; i++) {
      if (s2[i] != s2[i - 1]) {
        b.push_back(i);
      }
    }
    reverse(b.begin(), b.end());
    cout << (long long int)a.size() + (long long int)b.size() << " ";
    for (auto x : a) cout << x << " ";
    for (auto x : b) cout << x << " ";
    cout << "\n";
  }
  return 0;
}

### Prompt

Generate 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>
long long int inf = 1e18;
using namespace std;
long long int add(long long a, long long b) {
  return ((a % 1000000007) + (b % 1000000007)) % 1000000007;
}
long long int subtract(long long a, long long b) {
  return ((a % 1000000007) - (b % 1000000007)) % 1000000007;
}
long long int mult(long long a, long long b) {
  return ((a % 1000000007) * (b % 1000000007)) % 1000000007;
}
long long int add1(long long a, long long b) {
  return ((a % 998244353) + (b % 998244353)) % 998244353;
}
long long int subtract1(long long a, long long b) {
  return ((a % 998244353) - (b % 998244353)) % 998244353;
}
long long int mult1(long long a, long long b) {
  return ((a % 998244353) * (b % 998244353)) % 998244353;
}
long long int binmodwithMOD1(long long int a, long long int b) {
  long long int res = 1;
  while (b) {
    if (b & 1) res = mult1(res, a) % 998244353;
    a = mult1(a, a) % 998244353;
    b >>= 1;
  }
  return res;
}
long long int binmodwithMOD(long long int a, long long int b) {
  long long int res = 1;
  while (b) {
    if (b & 1) res = mult(res, a) % 1000000007;
    a = mult(a, a) % 1000000007;
    b >>= 1;
  }
  return res;
}
long long int binmodwithoutMOD1orMOD2(long long int a, long long int b) {
  long long int res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int inverse(long long int n) {
  return binmodwithMOD(n, 1000000007 - 2);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    string s1, s2;
    cin >> s1 >> s2;
    vector<long long int> a, b;
    for (long long int i = 1; i < n; i++) {
      if (s1[i] != s1[i - 1]) {
        a.push_back(i);
      }
    }
    if (s1[n - 1] != s2[n - 1]) a.push_back(n);
    for (long long int i = 1; i < n; i++) {
      if (s2[i] != s2[i - 1]) {
        b.push_back(i);
      }
    }
    reverse(b.begin(), b.end());
    cout << (long long int)a.size() + (long long int)b.size() << " ";
    for (auto x : a) cout << x << " ";
    for (auto x : b) cout << x << " ";
    cout << "\n";
  }
  return 0;
}

```