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;
const int M = 1000000007;
void _print(int a) { cerr << a; }
void _print(bool a) { cerr << a; }
void _print(long long a) { cerr << a; }
template <class T>
void _print(std::vector<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(std::set<T> s) {
  cerr << "[ ";
  for (T i : s) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(std::multiset<T> s) {
  cerr << "[ ";
  for (T i : s) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
vector<long long> sieve(int n) {
  int* arr = new int[n + 1]();
  vector<long long> vect;
  for (int i = 2; i <= n; i++)
    if (arr[i] == 0) {
      vect.push_back(i);
      for (int j = 2 * i; j <= n; j += i) arr[j] = 1;
    }
  return vect;
}
long long mod_add(long long a, long long b, long long M) {
  a = a % M;
  b = b % M;
  return (((a + b) % M) + M) % M;
}
long long mod_mul(long long a, long long b, long long M) {
  a = a % M;
  b = b % M;
  return (((a * b) % M) + M) % M;
}
long long mod_sub(long long a, long long b, long long M) {
  a = a % M;
  b = b % M;
  return (((a - b) % M) + M) % M;
}
int main() {
  long long t = 1;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    string a, b;
    cin >> a >> b;
    int r = (int)a.size();
    if (a == b) {
      cout << 0 << "\n";
      continue;
    }
    long long ans = 0;
    for (int i = 0; i < r; i++)
      if (a[i] != b[i]) ans++;
    cout << 3 * ans << " ";
    for (int i = r - 1; i >= 0; i--) {
      if (a[i] != b[i]) cout << i + 1 << " 1 " << i + 1 << " ";
    }
    cout << "\n";
  }
}