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;
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
void flip(string& a, int n) {
  for (int i = 0; i < n; i++) {
    if (a[i] == '0') {
      a[i] = '1';
    } else {
      a[i] = '0';
    }
  }
  reverse(a.begin(), a.begin() + n);
}
void solve() {
  int n;
  cin >> n;
  string a, b;
  cin >> a;
  cin >> b;
  vector<int> ans;
  int i = 0;
  int j = n - 1;
  int f = 0;
  int l = n - 1;
  while (j >= 0 && a[j] == b[j]) {
    j--;
    l--;
  }
  int flag = 0;
  while (j >= 0) {
    if (flag && a[f] == b[j]) {
      ans.push_back(j + 1);
      while (j >= 0 && a[f] == b[j]) {
        j--;
        f--;
        flag = 0;
      }
      int x = f;
      f = l;
      l = x;
    } else if (!flag && a[f] != b[j]) {
      ans.push_back(j + 1);
      while (j >= 0 && a[f] != b[j]) {
        j--;
        f++;
        flag = 1;
      }
      int x = f;
      f = l;
      l = x;
    } else {
      ans.push_back(1);
      if (a[f] == '0') {
        a[f] = '1';
      } else {
        a[f] = '0';
      }
    }
  }
  cout << ans.size() << " ";
  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i] << " ";
  }
  cout << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  cin >> t;
  while (t--) {
    solve();
  }
  cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC
       << "ms\n";
}
