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;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  for (const auto& x : v) os << x << " ";
  return os << "\n";
}
template <typename T>
istream& operator>>(istream& is, vector<T>& v) {
  for (auto& x : v) is >> x;
  return is;
}
ll power(ll a, ll n) {
  if (n == 0) {
    return 1;
  }
  ll sqrt_a = pow(a, n / 2);
  if (n % 2 == 1) {
    return sqrt_a * sqrt_a * a;
  } else {
    return sqrt_a * sqrt_a;
  }
}
template <typename T>
T sign(T a) {
  if (a < 0) return -1;
  if (a > 0) return 1;
  return 0;
}
template <typename T>
T log_2(T n) {
  T i = 1;
  T j = 0;
  while (i < n) {
    i *= 2;
    ++j;
  }
  return j;
}
template <typename T>
T mod(T a, T b) {
  T ans = a % b;
  return ans < 0 ? ans + b : ans;
}
ll gcd(ll a, ll b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
class Node {
 public:
  queue<pair<int, int>> q;
};
class Compare {
 public:
  bool operator()(pair<int, int>& a, pair<int, int>& b) {
    if (a.first < b.first) {
      return true;
    } else if (a.first == b.first) {
      return a.second < b.second;
    } else {
      return false;
    }
  }
};
vector<pair<int, int>> divs(int n) {
  vector<pair<int, int>> s;
  for (int i = 3; i <= sqrt(n); ++i) {
    if (n % i == 0) {
      if (i % 2 == 1 && (n / i) % 2 == 0) {
        s.emplace_back(i, n / i);
      }
      if (i % 2 == 0 && (n / i) % 2 == 1) {
        s.emplace_back(n / i, i);
      }
    }
  }
  return s;
}
ll MOD = 1000000007;
void find(vector<ll>& count, vector<int>& first, ll n) {
  if (n <= 4) {
    return;
  }
  ll ans = count[n - 2] + 2 * count[n - 3];
  if (first[n - 2] == 0 && first[n - 3] == 0) {
    ans += 4;
    first[n - 1] = 1;
  }
  count[n - 1] = ans % MOD;
}
void solve() {
  int n;
  cin >> n;
  auto a = vector<char>(n);
  auto b = vector<char>(n);
  auto ans = vector<int>();
  cin >> a >> b;
  for (int i = 0; i < n - 1; ++i) {
    if (a[0] == b[n - 1 - i]) {
      ans.push_back(1);
      a[0] = a[0] == '0' ? '1' : '0';
    }
    ans.push_back(n - i);
    for (int j = 0; j < n - i; ++j) {
      a[j] = a[j] == '0' ? '1' : '0';
    }
    for (int j = 0; j < (n - i) / 2; ++j) {
      swap(a[j], a[n - i - 1 - j]);
    }
  }
  if (a[0] != b[0]) {
    ans.push_back(1);
    a[0] = a[0] == '0' ? '1' : '0';
  }
  cout << ans.size() << " " << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ll t;
  cin >> t;
  while (t--) {
    solve();
  }
}