#include <bits/stdc++.h>
using namespace std;
void solve() {
  long long n;
  cin >> n;
  char c;
  long long a[n], b[n];
  for (long long i = 0; i < n; i++) {
    cin >> c;
    a[i] = c - '0';
  }
  for (long long i = 0; i < n; i++) {
    cin >> c;
    b[i] = c - '0';
  }
  long long f = 0, l = n - 1, sf = 0, sl = 0;
  long long k = 0;
  long long p[n * 2];
  for (long long i = 0; i < n - 1; i++) {
    if (((a[l] + sl) % 2) == b[n - 1 - i]) {
      if ((sl % 2) == 1) {
        l++;
      } else {
        l--;
      }
      continue;
    } else if ((a[l] + sl) % 2 != (a[f] + sl) % 2) {
      p[k++] = 1;
      sf++;
    }
    if ((a[l] + sl) % 2 != b[n - 1 - i]) {
      p[k++] = abs(l - f) + 1;
      long long tmp = l;
      if (sl % 2 == 0) {
        l = f + 1;
        f = tmp;
      } else {
        l = f - 1;
        f = tmp;
      }
      sl++;
      sf = 0;
    }
  }
  if ((a[l] + sl) % 2 != b[0]) {
    p[k++] = 1;
  }
  cout << k;
  for (long long i = 0; i < k; i++) {
    cout << ' ' << p[i];
  }
  cout << endl;
  return;
}
int32_t main() {
  long long t;
  cin >> t;
  while (t--) solve();
}

### 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>
using namespace std;
void solve() {
  long long n;
  cin >> n;
  char c;
  long long a[n], b[n];
  for (long long i = 0; i < n; i++) {
    cin >> c;
    a[i] = c - '0';
  }
  for (long long i = 0; i < n; i++) {
    cin >> c;
    b[i] = c - '0';
  }
  long long f = 0, l = n - 1, sf = 0, sl = 0;
  long long k = 0;
  long long p[n * 2];
  for (long long i = 0; i < n - 1; i++) {
    if (((a[l] + sl) % 2) == b[n - 1 - i]) {
      if ((sl % 2) == 1) {
        l++;
      } else {
        l--;
      }
      continue;
    } else if ((a[l] + sl) % 2 != (a[f] + sl) % 2) {
      p[k++] = 1;
      sf++;
    }
    if ((a[l] + sl) % 2 != b[n - 1 - i]) {
      p[k++] = abs(l - f) + 1;
      long long tmp = l;
      if (sl % 2 == 0) {
        l = f + 1;
        f = tmp;
      } else {
        l = f - 1;
        f = tmp;
      }
      sl++;
      sf = 0;
    }
  }
  if ((a[l] + sl) % 2 != b[0]) {
    p[k++] = 1;
  }
  cout << k;
  for (long long i = 0; i < k; i++) {
    cout << ' ' << p[i];
  }
  cout << endl;
  return;
}
int32_t main() {
  long long t;
  cin >> t;
  while (t--) solve();
}

```