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>
#pragma GCC optimize("O3", "unroll-loops")
#pragma GCC target("avx2")
long long int add(long long int x, long long int y) {
  long long int res = x + y;
  return (res >= 1000000007 ? res - 1000000007 : res);
}
long long int mul(long long int x, long long int y) {
  long long int res = x * y;
  return (res >= 1000000007 ? res % 1000000007 : res);
}
long long int sub(long long int x, long long int y) {
  long long int res = x - y;
  return (res < 0 ? res + 1000000007 : res);
}
long long int power(long long int x, long long int y) {
  long long int res = 1;
  x %= 1000000007;
  while (y) {
    if (y & 1) res = mul(res, x);
    y >>= 1;
    x = mul(x, x);
  }
  return res;
}
long long int mod_inv(long long int x) { return power(x, 1000000007 - 2); }
using namespace std;
char toggle(char a) {
  if (a == '1')
    return '0';
  else
    return '1';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    string a, b;
    int n;
    cin >> n;
    cin >> a >> b;
    vector<int> ans;
    for (int i = n - 1; i >= 0; i--) {
      if (a[i] == b[i])
        continue;
      else {
        if (a[0] == b[i]) {
          ans.push_back(1);
          a[0] = toggle(a[0]);
          ans.push_back(i + 1);
        } else {
          ans.push_back(i + 1);
        }
        for (int j = 0; j < i + 1; j++) a[j] = toggle(a[j]);
        reverse(a.begin(), a.begin() + i + 1);
      }
    }
    cout << ans.size() << " ";
    for (auto i : ans) cout << i << " ";
    cout << "\n";
  }
  cerr << "time taken : " << (float)clock() / CLOCKS_PER_SEC << " secs"
       << "\n";
  return 0;
}