#include <bits/stdc++.h>
using namespace std;
int n, m;
char a[200005], b[200005];
int rs, rt, l, r;
void rd() {
  cin >> n;
  scanf("%s", a + 1);
  scanf("%s", b + 1);
}
int gb() {
  if (rs)
    return (a[l] - '0') ^ rt;
  else
    return (a[r] - '0') ^ rt;
}
int gf() {
  if (!rs)
    return (a[l] - '0') ^ rt;
  else
    return (a[r] - '0') ^ rt;
}
int ans[200005], nn;
void wxw(int x) {
  nn++;
  ans[nn] = x;
}
signed main() {
  int T;
  cin >> T;
  while (T--) {
    rd();
    l = 1;
    r = n;
    rt = rs = 0;
    nn = 0;
    for (int i = n; i >= 1; i--) {
      int t = b[i] - '0';
      if (gb() == t) {
        if (rs)
          l++;
        else
          r--;
      } else {
        int u = gf();
        if (u == t) {
          wxw(1);
        }
        wxw(i);
        if (rs)
          r--;
        else
          l++;
        rs ^= 1;
        rt ^= 1;
      }
    }
    cout << nn << " ";
    for (int i = 1; i <= nn; i++) cout << ans[i] << " ";
    cout << "\n";
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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;
int n, m;
char a[200005], b[200005];
int rs, rt, l, r;
void rd() {
  cin >> n;
  scanf("%s", a + 1);
  scanf("%s", b + 1);
}
int gb() {
  if (rs)
    return (a[l] - '0') ^ rt;
  else
    return (a[r] - '0') ^ rt;
}
int gf() {
  if (!rs)
    return (a[l] - '0') ^ rt;
  else
    return (a[r] - '0') ^ rt;
}
int ans[200005], nn;
void wxw(int x) {
  nn++;
  ans[nn] = x;
}
signed main() {
  int T;
  cin >> T;
  while (T--) {
    rd();
    l = 1;
    r = n;
    rt = rs = 0;
    nn = 0;
    for (int i = n; i >= 1; i--) {
      int t = b[i] - '0';
      if (gb() == t) {
        if (rs)
          l++;
        else
          r--;
      } else {
        int u = gf();
        if (u == t) {
          wxw(1);
        }
        wxw(i);
        if (rs)
          r--;
        else
          l++;
        rs ^= 1;
        rt ^= 1;
      }
    }
    cout << nn << " ";
    for (int i = 1; i <= nn; i++) cout << ans[i] << " ";
    cout << "\n";
  }
  return 0;
}

```