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 gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int power(long long int x, long long int n, long long int m) {
  long long int res = 1;
  while (n > 0) {
    if (n & 1) res = (res * x) % m;
    x = (x * x) % m;
    n = n >> 1;
  }
  return res;
}
long long int modInverse(long long int n, long long int p) {
  return power(n, p - 2, p);
}
long long int nCr(long long int n, long long int r, long long int p) {
  if (r == 0) return 1;
  long long int fac[n + 1];
  fac[0] = 1;
  for (long long int i = 1; i <= n; i++) fac[i] = fac[i - 1] * i % p;
  return ((fac[n] * modInverse(fac[r], p) % p) % p * modInverse(fac[n - r], p) %
          p) %
         p;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    string s1, s2;
    cin >> s1 >> s2;
    vector<long long int> res;
    for (long long int i = n - 1; i >= 0; i--) {
      if (s1[i] != s2[i]) {
        if (s1[0] == s2[i]) {
          res.push_back(0);
          if (s1[0] == '0')
            s1[0] = '1';
          else
            s1[0] = '0';
          res.push_back(i);
          for (long long int j = 0; j <= i; j++) {
            if (s1[j] == '0')
              s1[j] = '1';
            else
              s1[j] = '0';
          }
          reverse(s1.begin(), s1.begin() + i + 1);
        } else {
          res.push_back(i);
          for (long long int j = 0; j <= i; j++) {
            if (s1[j] == '0')
              s1[j] = '1';
            else
              s1[j] = '0';
          }
          reverse(s1.begin(), s1.begin() + i + 1);
        }
      }
    }
    cout << res.size() << " ";
    for (long long int i = 0; i < res.size(); i++) cout << res[i] + 1 << " ";
    cout << endl;
  }
  return 0;
}