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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
void __print(int32_t x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  long long f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
char myHeap[200 * 1024 * 1024];
long long sz = 0;
void assert_tle(bool q) {
  while (!q)
    ;
}
void *operator new(std::size_t count) {
  sz += count;
  assert_tle(sz <= 200 * 1024 * 1024);
  return myHeap + sz - count;
}
void operator delete(void *ptr) {}
void fastIO() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
void solve() {
  long long n;
  cin >> n;
  string a, b;
  cin >> a >> b;
  vector<long long> ans;
  long long e = 0, o = n - 1;
  for (long long i = n - 1; i >= 0; i--) {
    long long turn = n - i - 1;
    long long cur;
    if (turn % 2 == 0) {
      cur = e;
      e++;
      if (a[cur] != b[i]) {
        ans.push_back(i + 1);
      } else {
        ans.push_back(1);
        ans.push_back(i + 1);
      }
    } else {
      cur = o;
      o--;
      if (a[cur] == b[i]) {
        ans.push_back(i + 1);
      } else {
        ans.push_back(1);
        ans.push_back(i + 1);
      }
    }
  }
  cout << ans.size() << " ";
  for (auto it : ans) cout << it << " ";
}
int32_t main() {
  fastIO();
  long long t;
  cin >> t;
  while (t--) {
    solve();
    cout << "\n";
  }
  return 0;
}