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;
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
bool isPrime(long long n) {
  long long c;
  if (n <= 1) return false;
  c = sqrt(n);
  for (int i = 2; i <= c; i++)
    if (n % i == 0) return false;
  return true;
}
char sw(char c) {
  if (c == '0') {
    return '1';
  } else {
    return '0';
  }
}
const long long N = 1e5 + 7;
long long l, k, d, tr, n, m, sum, mini, maxi, a, b, c, x, y;
long long t[N], t1[N];
set<int> vis;
vector<int> v;
string s, ss;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  for (cin >> tr; tr--;) {
    cin >> n >> s >> ss;
    a = 0;
    n--;
    while (n > 0) {
      if (ss[n] != s[n]) {
        if (s[n] != s[0]) {
          t[a] = 1;
          a++;
          if (s[0] == '0') {
            s[0] = '1';
          } else {
            s[0] = '0';
          }
        }
        for (int i = 0; i <= (n) / 2; i++) {
          if (s[i] == s[n - i]) {
            if (s[i] == '1') {
              s[i] = '0';
              s[n - i] = '0';
            } else {
              s[i] = '1';
              s[n - i] = '1';
            }
          }
        }
        t[a] = n + 1;
        a++;
      }
      n--;
    }
    if (s[0] != ss[0]) {
      s[0] = sw(s[0]);
      t[a] = 1;
      a++;
    }
    cout << a << endl;
    for (int i = 0; i < a; i++) {
      cout << t[i] << " ";
    }
    cout << endl;
  }
}
