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 lli = long long int;
using ulli = unsigned long long int;
using ld = long double;
template <class T>
using heap = priority_queue<T, vector<T>, greater<T>>;
template <class A, class B>
using hmap = unordered_map<A, B>;
template <class T>
using hset = unordered_set<T>;
template <class T>
struct Matrix : public vector<vector<T>> {
  Matrix(int dim1 = 0, int dim2 = 0, T v = T())
      : vector<vector<T>>(dim1, vector<T>(dim2, v)) {}
};
namespace std {
template <class T>
struct hash<vector<T>> {
  size_t operator()(vector<T> const& v) const {
    size_t c = v.size();
    for (auto& e : v) {
      c ^= (size_t)e + 0x9e3779b9 + (c << 6) + (c >> 2);
    }
    return c;
  }
};
}  // namespace std
const int INF = 1e9;
const lli LLINF = 4 * 1e18;
const lli MOD = 1e9 + 7;
const ld PI = 3.141592653589793;
template <typename T>
inline T sqr(T val) {
  return val * val;
}
template <class T>
inline void minset(T& a, T b) {
  if (b < a) a = b;
}
template <class T>
inline void maxset(T& a, T b) {
  if (a < b) a = b;
}
template <typename T>
T modul(T val, T __modulo = MOD) {
  return (val % __modulo + __modulo) % __modulo;
}
template <class V>
typename V::value_type sum(const V& els) {
  auto s = 0;
  for (const auto& e : els) {
    s += e;
  };
  return s;
}
template <class V>
typename V::value_type min(const V& els) {
  return *min_element(els.begin(), els.end());
}
template <class V>
typename V::value_type max(const V& els) {
  return *max_element(els.begin(), els.end());
}
struct __RM_UNUSED {
  __RM_UNUSED() {
    (void)INF;
    (void)LLINF;
    (void)PI;
  }
};
string tostr(char c) { return string(1, c); }
template <typename T>
string tostr(T c) {
  return to_string(c);
}
int toint(char c) { return c - '0'; }
int toint(string s, int base = 10) { return stoi(s, 0, base); }
template <class T>
inline T readget() {
  T v;
  cin >> v;
  return v;
}
template <class T>
istream& operator>>(istream& is, vector<T>& v) {
  for (T& e : v) {
    is >> e;
  }
  return is;
}
int main() {
  ios::sync_with_stdio(false);
  cout.tie(0);
  cin.tie(0);
  cout << fixed << setprecision(10);
  int nbTests;
  cin >> nbTests;
  for (int iTest = 0; iTest < nbTests; iTest++) {
    int n;
    cin >> n;
    vector<int> a(n), b(n);
    cin.ignore();
    for (int i = 0; i < n; i++) {
      char c;
      cin >> c;
      a[i] = c - '0';
    }
    cin.ignore();
    for (int i = 0; i < n; i++) {
      char c;
      cin >> c;
      b[i] = c - '0';
    }
    vector<int> r;
    for (int i = n - 1; i >= 0; i--) {
      if (a[i] != b[i]) {
        if (i == 0) {
          r.push_back(1);
        } else if (i == 1 || a[i - 1] == b[i - 1]) {
          r.push_back(i + 1);
          r.push_back(1);
          r.push_back(i + 1);
        } else if (a[i - 1] == a[i]) {
          r.push_back(i + 1);
          r.push_back(2);
          r.push_back(i + 1);
          i--;
        } else if (a[i - 2] == b[i - 2]) {
          r.push_back(i + 1);
          r.push_back(1);
          r.push_back(2);
          r.push_back(1);
          r.push_back(i + 1);
          i--;
        } else if (a[i - 1] == a[i - 2]) {
          r.push_back(i + 1);
          r.push_back(1);
          r.push_back(3);
          r.push_back(1);
          r.push_back(i + 1);
          i--;
          i--;
        } else {
          r.push_back(i + 1);
          r.push_back(3);
          r.push_back(i + 1);
          i--;
          i--;
        }
      }
    }
    cout << r.size();
    for (auto& e : r) {
      cout << " " << e;
    }
    cout << "\n";
  }
}