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.
#include <bits/stdc++.h>
using ll = long long;
using ull = unsigned long long;
using namespace std;
template <typename T, typename A>
int arg_max(std::vector<T, A> const& vec) {
  return static_cast<int>(
      std::distance(vec.begin(), max_element(vec.begin(), vec.end())));
}
template <typename T, typename A>
int arg_min(std::vector<T, A> const& vec) {
  return static_cast<int>(
      std::distance(vec.begin(), min_element(vec.begin(), vec.end())));
}
int dfs(const vector<vector<int>>& g, vector<int>& dist, int start,
        int end = -1) {
  queue<int> q;
  q.push(start);
  queue<int> d;
  d.push(0);
  while (q.size() > 0) {
    auto x = q.front();
    auto dd = d.front();
    q.pop();
    d.pop();
    if (dist[x] == std::numeric_limits<int>::max()) {
      dist[x] = dd;
      if (end == x) break;
      for (auto e : g[x]) {
        q.push(e);
        d.push(dd + 1);
      }
    }
  }
  if (end >= 0 && end < dist.size())
    return dist[end];
  else
    return -1;
}
ll largestPower(ll n, ll p) {
  ll x = 0;
  while (n) {
    n /= p;
    x += n;
  }
  return x;
}
ll power(ll x, ll y, ll p) {
  ll res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
ll modFact(ll n, ll p) {
  if (n >= p) return 0;
  ll res = 1;
  bool isPrime[n + 1];
  memset(isPrime, 1, sizeof(isPrime));
  for (ll i = 2; i * i <= n; i++) {
    if (isPrime[i]) {
      for (ll j = 2 * i; j <= n; j += i) isPrime[j] = 0;
    }
  }
  for (ll i = 2; i <= n; i++) {
    if (isPrime[i]) {
      ll k = largestPower(n, i);
      res = (res * power(i, k, p)) % p;
    }
  }
  return res;
}
void invert(char& x) {
  if (x == '1')
    x = '0';
  else
    x = '1';
}
int main() {
  ll tt = 1;
  cin >> tt;
  for (ll t = 0; t < tt; ++t) {
    ll n;
    cin >> n;
    string a, b;
    cin >> a >> b;
    vector<int> ops;
    for (int i = n - 1; i > 0; --i) {
      if (a[i] == b[i])
        continue;
      else {
        if (a[0] == b[i]) {
          ops.push_back(1);
          invert(a[0]);
        }
        ops.push_back(i + 1);
        for (ll j = 0; j < i + 1; ++j) invert(a[j]);
        reverse(a.begin(), a.begin() + i + 1);
      }
    }
    if (a[0] != b[0]) ops.push_back(1);
    cout << ops.size() << " ";
    for (auto o : ops) cout << o << " ";
    cout << "\n";
  }
  return 0;
}
