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>
const int N = 200002;
int T, i, n, cnt, op, p[N], t;
char a[N], b[N];
void calc(int k, int l, int r) {
  if (k == 0) return;
  if (b[k] == (a[r] ^ t)) {
    if (l < r)
      calc(k - 1, l, r - 1);
    else
      calc(k - 1, l, r + 1);
  } else {
    if (a[l] != a[r]) p[++cnt] = 1;
    p[++cnt] = k;
    t ^= 1;
    if (l < r)
      calc(k - 1, r, l + 1);
    else
      calc(k - 1, r, l - 1);
  }
}
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%d%s%s", &n, a + 1, b + 1);
    cnt = 0;
    t = 0;
    calc(n, 1, n);
    printf("%d ", cnt);
    for (i = 1; i <= cnt; i++) printf("%d ", p[i]);
    printf("\n");
  }
}