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;
const int N = 1e6 + 5;
int _w;
char A[N], B[N];
int n, ans[N], res, l, r, t;
int get(int p) {
  if (l <= r)
    return A[l + p - 1] ^ t;
  else
    return A[l - p + 1] ^ t;
}
inline void solve(void) {
  _w = scanf("%d%s%s", &n, A + 1, B + 1), res = 0;
  l = 1, r = n, t = 0;
  for (int i = n; i; --i) {
    if (get(i) ^ B[i]) {
      if ((A[l] ^ t) == B[i]) ans[++res] = 1;
      swap(l, r), t ^= 1, ans[++res] = i;
    }
    if (r > l)
      --r;
    else
      ++r;
  }
  printf("%d ", res);
  for (int i = 1; i <= res; ++i) printf("%d ", ans[i]);
  puts("");
}
int main(void) {
  int T;
  _w = scanf("%d", &T);
  while (T--) solve();
  return 0;
}