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;
long long l, r, a, x, y, z, b, i, j, k, n, m, num, ans, t;
long long arr[200005], arr1[200005];
char s[100005], str[100005];
bool cmp(long long a, long long b) { return a > b; }
int main() {
  cin >> t;
  for (k = 0; k < t; k++) {
    cin >> n;
    cin >> s >> str;
    num = 0;
    j = 0;
    long long flag = 0;
    if (s[0] == str[n - 1]) {
      arr[j++] = 1;
      arr[j++] = n;
    } else
      arr[j++] = n;
    num++;
    long long h = 1;
    arr1[0] = n - 1;
    arr1[1] = 1;
    for (i = 2; i < n - 1; i++) {
      if (i % 2 == 0)
        arr1[i] = arr1[i - 2] - 1;
      else
        arr1[i] = arr1[i - 2] + 1;
    }
    for (i = 0; i < n - 1; i++) {
      if (num % 2 == 1) {
        if (s[arr1[i]] == '1')
          s[arr1[i]] = '0';
        else
          s[arr1[i]] = '1';
      }
      if (s[arr1[i]] == str[n - h - 1]) {
        arr[j++] = 1;
        arr[j++] = n - h;
      } else
        arr[j++] = n - h;
      num++;
      h++;
    }
    cout << j << ' ';
    for (i = 0; i < j; i++) {
      cout << arr[i] << ' ';
    }
    cout << '\n';
  }
  return 0;
}