This is the easy 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 3n 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≤ 1000) — 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 1000.

Output

For each test case, output an integer k (0≤ k≤ 3n), 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.
#include <bits/stdc++.h>
using namespace std;
int arr[100005][2];
char str[100005];
vector<int> vec;
int main() {
  int i, j, k, a, b, c, n, m, t, T, cnt;
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &n);
    vec.clear();
    for (i = 0; i < 2; i++) {
      scanf("%s", str);
      for (j = 0; j < n; j++) arr[j + 1][i] = str[j] - '0';
    }
    m = 0;
    cnt = 0;
    for (i = n; i >= 2; i--) {
      if (cnt % 2 == 0)
        j = i - m;
      else
        j = -i + m;
      if ((arr[j][0] + cnt) % 2 == arr[i][1]) continue;
      if (arr[i][1] == arr[1][0]) {
        vec.push_back(1);
        arr[1][0] = 1 - arr[1][0];
      }
      vec.push_back(i);
      m = 1 + i - m;
      arr[1][0] = 1 - (arr[j][0] + cnt) % 2;
      cnt++;
    }
    if (arr[1][0] != arr[1][1]) vec.push_back(1);
    printf("%d ", vec.size());
    for (i = 0; i < vec.size(); i++) printf("%d ", vec[i]);
    printf("\n");
  }
}
