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 namespace std;
bool isPrime[100005] = {false};
void sieve() {
  int i;
  for (i = 2; i < 100005; ++i) isPrime[i] = 1;
  isPrime[0] = isPrime[1] = 0;
  for (i = 2; i * i < 100005; ++i) {
    if (isPrime[i]) {
      for (int j = i * i; j < 100005; j += i) isPrime[j] = 0;
    }
  }
}
long long fastPower(long long a, long long b) {
  if (b == 0) return 1;
  long long prod = 1;
  while (b > 0) {
    if (b & 1) prod = prod * a;
    b = b >> 1;
    a = a * a;
  }
  return prod;
}
bool flip(bool f) {
  if (f) return 0;
  return 1;
}
char flipchr(char a) {
  if (a == '1') return '0';
  return '1';
}
void printa(bool f) {
  if (f)
    cout << "YES"
         << "\n";
  else
    cout << "NO"
         << "\n";
}
void getAns(string &a, int l, int r) {
  int n = a.length(), i, j;
  for (i = l; i < r + 1; ++i) {
    a[i] = flipchr(a[i]);
  }
  reverse(a.begin(), a.begin() + r + 1);
}
void solve() {
  int n, i;
  cin >> n;
  string a, b;
  cin >> a >> b;
  vector<int> ans;
  i = n - 1;
  while (i >= 0) {
    if (a[i] != b[i]) {
      if (a[0] == b[i]) {
        ans.push_back(1);
        a[0] = flipchr(a[0]);
      }
      getAns(a, 0, i);
      ans.push_back(i + 1);
    }
    --i;
  }
  cout << ans.size() << ' ';
  for (auto t : ans) cout << t << ' ';
  cout << "\n";
  ;
}
int main() {
  int tc = 1;
  scanf("%d", &tc);
  while (tc--) {
    solve();
  }
  return 0;
}
