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 MAXN = 1e5 + 7;
int n, T, tot;
int ans[MAXN], a[MAXN];
char s[MAXN], t[MAXN];
inline void Br(int L, int R) {
  for (int i = L; i <= R; ++i) a[i] ^= 1;
  for (int i = L; i <= R; ++i) {
    if (R - i + L <= i) break;
    swap(a[i], a[R - i + L]);
  }
  return;
}
int main() {
  scanf("%d", &T);
  while (T-- > 0) {
    tot = 0;
    scanf("%d", &n);
    scanf("%s", s + 1);
    scanf("%s", t + 1);
    for (int i = 1; i <= n; ++i) {
      a[i] = (s[i] - '0');
    }
    for (int i = n; i >= 1; --i) {
      if (a[i] == t[i] - '0') continue;
      int pos = -1;
      for (int j = i; j >= 1; --j) {
        if (a[j] == t[i] - '0') {
          pos = j;
          break;
        }
      }
      if (pos == -1) {
        Br(1, i);
        ans[++tot] = i;
      } else {
        Br(1, pos);
        ans[++tot] = pos;
        Br(1, i);
        ans[++tot] = i;
      }
    }
    printf("%d ", tot);
    for (int i = 1; i <= tot; ++i) printf("%d ", ans[i]);
    puts("");
  }
  return 0;
}