#include <bits/stdc++.h>
using namespace std;
int n;
char _a[200020], _b[200020];
int ans[200020 << 2], a[200020], b[200020], cs = 0;
bool ok = 0;
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &n);
    scanf("%s%s", _a + 1, _b + 1);
    for (register int i = 1; i <= n; i++)
      a[i] = _a[i] - '0', b[i] = _b[i] - '0';
    int l = 1, r = n, ok = 0;
    for (register int i = n; i >= 1; i--) {
      int x;
      if (l < r)
        x = a[i - 1 + l];
      else
        x = a[l - i + 1];
      x ^= ok;
      if (x == b[i]) {
        if (l < r)
          r--;
        else
          r++;
        continue;
      }
      x = a[l] ^ ok;
      if ((x ^ 1) != b[i]) {
        ans[++cs] = 1;
        a[l] ^= 1;
      }
      ans[++cs] = i;
      swap(l, r);
      ok ^= 1;
      if (l < r)
        r--;
      else
        r++;
    }
    printf("%d", cs);
    for (register int i = 1; i <= cs; i++) printf(" %d", ans[i]);
    printf("\n");
    cs = 0;
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
char _a[200020], _b[200020];
int ans[200020 << 2], a[200020], b[200020], cs = 0;
bool ok = 0;
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &n);
    scanf("%s%s", _a + 1, _b + 1);
    for (register int i = 1; i <= n; i++)
      a[i] = _a[i] - '0', b[i] = _b[i] - '0';
    int l = 1, r = n, ok = 0;
    for (register int i = n; i >= 1; i--) {
      int x;
      if (l < r)
        x = a[i - 1 + l];
      else
        x = a[l - i + 1];
      x ^= ok;
      if (x == b[i]) {
        if (l < r)
          r--;
        else
          r++;
        continue;
      }
      x = a[l] ^ ok;
      if ((x ^ 1) != b[i]) {
        ans[++cs] = 1;
        a[l] ^= 1;
      }
      ans[++cs] = i;
      swap(l, r);
      ok ^= 1;
      if (l < r)
        r--;
      else
        r++;
    }
    printf("%d", cs);
    for (register int i = 1; i <= cs; i++) printf(" %d", ans[i]);
    printf("\n");
    cs = 0;
  }
  return 0;
}

```