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;
const int N = 100005;
char s[N], t[N];
int num[N], des[N];
vector<int> ans;
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    int n;
    scanf("%d %s %s", &n, s + 1, t + 1);
    if (n == 1) {
      if (s[1] != t[1]) {
        puts("1 1");
      } else {
        puts("0");
      }
      continue;
    }
    for (int i = (1); i <= (n); ++i) num[i] = s[i] - '0', des[i] = t[i] - '0';
    int bit = num[1];
    for (int i = (2); i <= (n); ++i) {
      if (bit != num[i]) {
        ans.push_back(i - 1);
      }
      bit = num[i];
    }
    for (int i = (n); i >= (1); --i) {
      if (bit != des[i]) {
        ans.push_back(i);
      }
      bit = des[i];
    }
    printf("%d ", (int)(ans).size());
    for (int e : ans) printf("%d ", e);
    puts("");
    ans.clear();
  }
  return 0;
}
