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 maxl = 3e5 + 10;
int n, m, cas, k, cnt, tot;
int ans[maxl];
int a[maxl], b[maxl], tmp[maxl];
char s[maxl];
bool in[maxl];
inline void prework() {
  scanf("%d", &n);
  scanf("%s", s + 1);
  for (int i = 1; i <= n; i++) a[i] = s[i] - '0';
  scanf("%s", s + 1);
  for (int i = 1; i <= n; i++) b[i] = s[i] - '0';
}
inline void mainwork() {
  ans[0] = 0;
  for (int i = n; i >= 2; i--)
    if (a[i] != b[i]) {
      if (a[1] == b[i]) ans[++ans[0]] = 1, a[1] ^= 1;
      ans[++ans[0]] = i;
      for (int j = 1; j <= i; j++) tmp[j] = a[i - j + 1] ^ 1;
      for (int j = 1; j <= i; j++) a[j] = tmp[j];
    }
  if (a[1] != b[1]) ans[++ans[0]] = 1;
}
inline void print() {
  printf("%d ", ans[0]);
  for (int i = 1; i <= ans[0]; i++) printf("%d ", ans[i]);
  puts("");
}
int main() {
  int t = 1;
  scanf("%d", &t);
  for (cas = 1; cas <= t; cas++) {
    prework();
    mainwork();
    print();
  }
  return 0;
}