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;
namespace io {
inline char nc() {
  static char buf[100000], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
template <class I>
inline void fr(I &num) {
  num = 0;
  register char c = nc();
  register int p = 1;
  while (c < '0' || c > '9') c == '-' ? p = -1, c = nc() : c = nc();
  while (c >= '0' && c <= '9') num = num * 10 + c - '0', c = nc();
  num *= p;
}
};  // namespace io
using io ::fr;
const int N = 1000005;
char s[N], t[N];
int n, T, ans[N];
int main() {
  scanf("%d", &T);
  register int i, j, k, cn;
  while (T--) {
    scanf("%d%s%s", &n, s + 1, t + 1);
    for (i = n, j = 1, k = cn = 0; n; ++cn, ans[++k] = n--)
      cn & 1 ? (s[i--] != t[n] ? ans[++k] = 1 : 0)
             : (s[j++] == t[n] ? ans[++k] = 1 : 0);
    printf("%d ", k);
    for (i = 1; i <= k; ++i) printf("%d ", ans[i]);
    puts("");
  }
  return 0;
}