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 int c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, t, x, y, z;
long long int mn = 8675435789, mx = -8675868757, PI = 3.14159265359,
              md = 1000000007, aa = 0, bb = 0, cc = 0;
string r, s, u, v, w;
long long int a[250002], b[250002];
map<long long int, long long int> mp, mq;
queue<long long int> qu, qe;
pair<long long int, long long int> pp[30030];
set<string> se, sa;
int main() {
  {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
  }
  cin >> t;
  while (t--) {
    cin >> n >> s >> r;
    cc = 0;
    for (i = n - 1; i >= 0; i--) {
      if (s[i] != r[i]) {
        if (s[0] == r[i]) {
          a[cc] = 1;
          cc++;
          if (s[0] == '1')
            s[0] = '0';
          else
            s[0] = '1';
        }
        a[cc] = i + 1;
        cc++;
        for (j = 0; j < i + 1; j++) {
          if (s[j] == '1')
            s[j] = '0';
          else
            s[j] = '1';
        }
        reverse(s.begin(), s.begin() + i + 1);
      }
    }
    cout << cc << ' ';
    for (i = 0; i < cc; i++) cout << a[i] << ' ';
    cout << endl;
  }
}