Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int tc, n;
char A[100005], B[100005];
vector<int> ans;
int main() {
  scanf("%d", &tc);
  while (tc--) {
    scanf("%d", &n);
    scanf("%s", &A);
    scanf("%s", &B);
    ans.clear();
    int inv = 0;
    int l = 0, r = n - 1;
    for (int i = n - 1; i >= 0; i--) {
      if (!inv) {
        if (A[r] == B[i]) {
          r--;
          continue;
        }
        if (A[l] == B[i]) {
          ans.push_back(1);
          ans.push_back(i + 1);
          l++;
          inv ^= 1;
        } else {
          ans.push_back(i + 1);
          l++;
          inv ^= 1;
        }
      } else {
        if (A[l] != B[i]) {
          l++;
          continue;
        }
        if (A[r] != B[i]) {
          ans.push_back(1);
          ans.push_back(i + 1);
          r--;
          inv ^= 1;
        } else {
          ans.push_back(i + 1);
          r--;
          inv ^= 1;
        }
      }
    }
    if ((int)(ans.size()) > 2 * n) assert(1 == 2);
    printf("%d", (int)(ans.size()));
    for (int v : ans) {
      printf(" %d", v);
    }
    printf("\n");
  }
  return 0;
}