#include <bits/stdc++.h>
const int MAXN = 1e5 + 10;
const int mod = 1e9 + 7;
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int inf = 0x3f3f3f3f;
using namespace std;
char s1[MAXN], s2[MAXN];
int tot[MAXN * 2], to;
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n;
    to = 0;
    scanf("%d", &n);
    scanf("%s", s1 + 1);
    scanf("%s", s2 + 1);
    int ans = 0;
    int need;
    for (need = n; need >= 1; need--) {
      if (s1[need] != s2[need]) {
        break;
      }
    }
    int l = 1, r = need, flag = 1;
    while (need) {
      if (flag) {
        if (s1[l] != s2[need]) {
          tot[to++] = need;
        } else {
          tot[to++] = 1;
          tot[to++] = need;
        }
        l++;
      } else {
        char x = s1[r];
        if (x == '0')
          x = '1';
        else
          x = '0';
        if (x != s2[need]) {
          tot[to++] = need;
        } else {
          tot[to++] = 1;
          tot[to++] = need;
        }
        r--;
      }
      flag ^= 1;
      need--;
    }
    printf("%d ", to);
    for (int i = 0; i < to; i++) printf("%d ", tot[i]);
    printf("\n");
  }
  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>
const int MAXN = 1e5 + 10;
const int mod = 1e9 + 7;
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int inf = 0x3f3f3f3f;
using namespace std;
char s1[MAXN], s2[MAXN];
int tot[MAXN * 2], to;
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n;
    to = 0;
    scanf("%d", &n);
    scanf("%s", s1 + 1);
    scanf("%s", s2 + 1);
    int ans = 0;
    int need;
    for (need = n; need >= 1; need--) {
      if (s1[need] != s2[need]) {
        break;
      }
    }
    int l = 1, r = need, flag = 1;
    while (need) {
      if (flag) {
        if (s1[l] != s2[need]) {
          tot[to++] = need;
        } else {
          tot[to++] = 1;
          tot[to++] = need;
        }
        l++;
      } else {
        char x = s1[r];
        if (x == '0')
          x = '1';
        else
          x = '0';
        if (x != s2[need]) {
          tot[to++] = need;
        } else {
          tot[to++] = 1;
          tot[to++] = need;
        }
        r--;
      }
      flag ^= 1;
      need--;
    }
    printf("%d ", to);
    for (int i = 0; i < to; i++) printf("%d ", tot[i]);
    printf("\n");
  }
  return 0;
}

```