#include <bits/stdc++.h>
using namespace std;
long long int inf = 1e18;
long long int p = 998244353;
long long int phi(long long int n) {
  long long int result = n;
  for (long long int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      while (n % i == 0) n /= i;
      result -= result / i;
    }
  }
  if (n > 1) result -= result / n;
  return result;
}
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long int modInverse(long long int n, long long int p) {
  return power(n, p - 2, p);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int i, j, y, x, k, z, w, g, m, a, b, c, key, q, l, r, n, t;
  long long int t2, t3, t4, t1;
  string s, u;
  cin >> t;
  while (t--) {
    cin >> n;
    cin >> s >> u;
    vector<long long int> ans;
    long long int ctr = 0;
    for (i = 0; i < n; i++) {
      if (s[i] == u[i]) {
        if (ctr == 1) {
          ctr = 0;
          if (i != 1) {
            ans.push_back(i);
            ans.push_back(1);
            ans.push_back(i);
          } else
            ans.push_back(1);
        }
      } else {
        if (ctr == 0)
          ctr++;
        else {
          ctr = 0;
          if (s[i] != s[i - 1]) {
            ctr = 0;
            ans.push_back(i);
            ans.push_back(i + 1);
            ans.push_back(i);
            if (i - 1 > 0) ans.push_back(i - 1);
          } else {
            if (i != 1) {
              ans.push_back(i + 1);
              ans.push_back(2);
              ans.push_back(i + 1);
            } else
              ans.push_back(2);
          }
        }
      }
    }
    if (ctr == 1) {
      ctr = 0;
      if (i != 1) {
        ans.push_back(i);
        ans.push_back(1);
        ans.push_back(i);
      } else
        ans.push_back(1);
    }
    cout << ans.size() << " ";
    for (auto nm : ans) cout << nm << " ";
    cout << "\n";
  }
  return 0;
}

### Prompt

Develop a solution in cpp 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;
long long int inf = 1e18;
long long int p = 998244353;
long long int phi(long long int n) {
  long long int result = n;
  for (long long int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      while (n % i == 0) n /= i;
      result -= result / i;
    }
  }
  if (n > 1) result -= result / n;
  return result;
}
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long int modInverse(long long int n, long long int p) {
  return power(n, p - 2, p);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int i, j, y, x, k, z, w, g, m, a, b, c, key, q, l, r, n, t;
  long long int t2, t3, t4, t1;
  string s, u;
  cin >> t;
  while (t--) {
    cin >> n;
    cin >> s >> u;
    vector<long long int> ans;
    long long int ctr = 0;
    for (i = 0; i < n; i++) {
      if (s[i] == u[i]) {
        if (ctr == 1) {
          ctr = 0;
          if (i != 1) {
            ans.push_back(i);
            ans.push_back(1);
            ans.push_back(i);
          } else
            ans.push_back(1);
        }
      } else {
        if (ctr == 0)
          ctr++;
        else {
          ctr = 0;
          if (s[i] != s[i - 1]) {
            ctr = 0;
            ans.push_back(i);
            ans.push_back(i + 1);
            ans.push_back(i);
            if (i - 1 > 0) ans.push_back(i - 1);
          } else {
            if (i != 1) {
              ans.push_back(i + 1);
              ans.push_back(2);
              ans.push_back(i + 1);
            } else
              ans.push_back(2);
          }
        }
      }
    }
    if (ctr == 1) {
      ctr = 0;
      if (i != 1) {
        ans.push_back(i);
        ans.push_back(1);
        ans.push_back(i);
      } else
        ans.push_back(1);
    }
    cout << ans.size() << " ";
    for (auto nm : ans) cout << nm << " ";
    cout << "\n";
  }
  return 0;
}

```