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>
#pragma optimize("O3")
#pragma GCC target("sse4")
using namespace std;
template <int D, typename T>
struct Vec : public vector<Vec<D - 1, T>> {
  static_assert(D >= 1, "Vector dimension must be greater than zero!");
  template <typename... Args>
  Vec(int n = 0, Args... args)
      : vector<Vec<D - 1, T>>(n, Vec<D - 1, T>(args...)) {}
};
template <typename T>
struct Vec<1, T> : public vector<T> {
  Vec(int n = 0, const T& val = T()) : vector<T>(n, val) {}
};
long long int modpower(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  if (y == 0) return 1;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long int power(long long int x, long long int y) {
  long long int res = 1;
  x = x;
  if (y == 0) return 1;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
long long int modinv(long long int a, long long int p) {
  return modpower(a, p - 2, p);
}
const int dx[8] = {1, 0, -1, 0, 1, 1, -1, -1},
          dy[8] = {0, 1, 0, -1, -1, 1, -1, 1};
long long safe_sqrt(long long n) {
  long long res = sqrt((long double)n);
  if (res * res == n) return res;
  const long double eps = 1e-8;
  if ((long double)(res + 1) * (res + 1) < n + eps) return res + 1;
  if ((long double)res * res + eps > n) return res - 1;
  return res;
}
long long int n;
string a, b;
void fun(long long int idx = 0) {
  vector<char> v(idx + 1);
  for (long long int i = 0; i <= idx; i += 1) {
    v[i] = '1' - a[i] + '0';
  }
  for (long long int i = 0; i <= idx; i += 1) {
    a[i] = v[idx - i];
  }
  return;
}
void solve() {
  cin >> n;
  a = "";
  b = "";
  cin >> a >> b;
  vector<long long int> ans;
  char a_0 = a[0];
  long long int cnt = 0;
  long long int val = 0;
  for (long long int i = n - 1; i >= 0; i--) {
    char a_i;
    long long int idx;
    if (cnt % 2 == 0) {
      idx = val + i;
      a_i = a[idx];
    } else {
      idx = val - i;
      a_i = '1' - a[idx] + '0';
    }
    if (a_i != b[i]) {
      if (a_0 != b[i]) {
        ans.push_back(i + 1);
        a_0 = '1' - a_i + '0';
      } else {
        a_0 = '1' - a_0 + '0';
        ans.push_back(1);
        ans.push_back(i + 1);
        a_0 = '1' - a_i + '0';
      }
      if (cnt % 2 == 0) {
        val += i;
      } else {
        val -= i;
      }
      cnt++;
    }
  }
  cout << ans.size() << '\n';
  for (auto v : ans) {
    cout << v << " ";
  }
  cout << '\n';
  return;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int T = 1;
  cin >> T;
  for (long long int c = 0; c < T; c++) {
    solve();
  }
  return 0;
}
