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;
inline int read() {
  char c = getchar();
  int x = 0, f = 1;
  while ((c < '0' || c > '9') && c ^ '-') c = getchar();
  if (c == '-') f = -1, c = getchar();
  while (c >= '0' && c <= '9') {
    x = (x << 3) + (x << 1) + c - 48;
    c = getchar();
  }
  return x * f;
}
const int N = 1e3 + 10;
char c1[N], c2[N];
int n, t, a[N], tmp[N], top = 0, dan[N << 2], b[N];
inline void caozuo(int x) {
  dan[++top] = x;
  for (int i = 1; i <= x; i++) tmp[i] = a[i] ^ 1;
  for (int i = 1; i <= x; i++) a[i] = tmp[x - i + 1];
}
int main() {
  t = read();
  while (t--) {
    n = read();
    top = 0;
    scanf("%s %s", c1 + 1, c2 + 1);
    for (int i = 1; i <= n; i++) {
      a[i] = c1[i] - 48;
      b[i] = c2[i] - 48;
    }
    for (int i = 1; i <= n; i++) {
      if (a[1] == b[n - i + 1]) caozuo(1);
      caozuo(n - i + 1);
    }
    cout << top << " ";
    for (int i = 1; i <= top; i++) cout << dan[i] << " ";
    cout << "\n";
  }
  return 0;
}
