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;
template <typename T>
void inpos(T &x) {
  x = 0;
  register T c = getchar();
  while (((c < 48) || (c > 57)) && (c != '-')) c = getchar();
  bool neg = 0;
  if (c == '-') neg = 1;
  for (; c < 48 || c > 57; c = getchar())
    ;
  for (; c > 47 && c < 58; c = getchar()) x = (x << 3) + (x << 1) + (c & 15);
  if (neg) x = -x;
}
template <typename T>
void outpos(T n) {
  if (n < 0) {
    putchar('-');
    n *= -1;
  }
  char snum[65];
  long long i = 0;
  do {
    snum[i++] = n % 10 + '0';
    n /= 10;
  } while (n);
  i = i - 1;
  while (i >= 0) putchar(snum[i--]);
}
inline void instr(char *str) {
  register char c = 0;
  register long long i = 0;
  while (c < 33) c = getchar();
  while (c != '\n' && c != ' ' && c != EOF) {
    str[i] = c;
    c = getchar();
    ++i;
  }
  str[i] = '\0';
}
template <typename T, typename TT>
inline void inpos(T &n, TT &m) {
  inpos(n);
  inpos(m);
}
template <typename T, typename TT, typename TTT>
inline void inpos(T &n, TT &m, TTT &o) {
  inpos(n, m);
  inpos(o);
}
template <typename T, typename TT, typename TTT, typename TTTT>
inline void inpos(T &n, TT &m, TTT &o, TTTT &p) {
  inpos(n, m);
  inpos(o, p);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    string A;
    string B;
    cin >> A;
    cin >> B;
    deque<char> As;
    deque<char> Bs;
    for (auto elem : A) {
      As.push_back(elem);
    }
    for (auto elem : B) {
      Bs.push_back(elem);
    }
    bool flipped = false;
    bool moov = false;
    vector<long long> ans;
    for (long long j = n - 1; j >= 1; j--) {
      if (As.back() == Bs.back()) {
        As.pop_back();
        Bs.pop_back();
      } else {
        if (As.front() == As.back()) {
          ans.push_back(j + 1);
          ans.push_back(j);
          As.pop_front();
          Bs.pop_back();
        } else {
          ans.push_back(1);
          ans.push_back(j + 1);
          ans.push_back(j);
          As.pop_front();
          Bs.pop_back();
        }
      }
    }
    if (As.back() != Bs.back()) {
      ans.push_back(1);
    }
    cout << ans.size() << " ";
    for (auto elem : ans) {
      cout << elem << " ";
    }
    cout << '\n';
  }
}