#include <bits/stdc++.h>
int dr[] = {2, 2, -2, -2, 1, -1, 1, -1};
int dc[] = {1, -1, 1, -1, 2, 2, -2, -2};
int dr1[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dc1[] = {1, -1, 1, 0, -1, 0, 1, -1};
int dr2[] = {0, 0, 1, -1};
int dc2[] = {1, -1, 0, 0};
using namespace std;
long long a[2005 * 2];
long long dp[2005 * 2][2005][2], vis[2005 * 2][2005][2], t, n;
long long b[2005 * 2];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long tc, i, j, cas = 0, m;
  cin >> tc;
  while (tc--) {
    cin >> n;
    string t, t1;
    cin >> t >> t1;
    long long f = 0;
    long long pre = 0, suf = n - 1;
    vector<long long> v;
    for (i = n - 1; i >= 0; i--) {
      if (f == 0) {
        if (t[pre] == t1[i]) v.push_back(0);
        v.push_back(i);
        pre++;
        f ^= 1;
      } else {
        if (t[suf] != t1[i]) v.push_back(0);
        v.push_back(i);
        suf--;
        f ^= 1;
      }
    }
    cout << v.size() << " ";
    for (auto x : v) cout << x + 1 << " ";
    cout << endl;
  }
  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>
int dr[] = {2, 2, -2, -2, 1, -1, 1, -1};
int dc[] = {1, -1, 1, -1, 2, 2, -2, -2};
int dr1[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dc1[] = {1, -1, 1, 0, -1, 0, 1, -1};
int dr2[] = {0, 0, 1, -1};
int dc2[] = {1, -1, 0, 0};
using namespace std;
long long a[2005 * 2];
long long dp[2005 * 2][2005][2], vis[2005 * 2][2005][2], t, n;
long long b[2005 * 2];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long tc, i, j, cas = 0, m;
  cin >> tc;
  while (tc--) {
    cin >> n;
    string t, t1;
    cin >> t >> t1;
    long long f = 0;
    long long pre = 0, suf = n - 1;
    vector<long long> v;
    for (i = n - 1; i >= 0; i--) {
      if (f == 0) {
        if (t[pre] == t1[i]) v.push_back(0);
        v.push_back(i);
        pre++;
        f ^= 1;
      } else {
        if (t[suf] != t1[i]) v.push_back(0);
        v.push_back(i);
        suf--;
        f ^= 1;
      }
    }
    cout << v.size() << " ";
    for (auto x : v) cout << x + 1 << " ";
    cout << endl;
  }
  return 0;
}

```