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;
const int dr[]{-1, -1, 0, 1, 1, 1, 0, -1};
const int dc[]{0, 1, 1, 1, 0, -1, -1, -1};
void run() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
bool valid(string &a, string &b, vector<int> out) {
  for (auto &idx : out) {
    for (int i = 0; i < idx; i++) a[i] ^= 1;
    reverse(a.begin(), a.begin() + idx);
  }
  return a == b;
}
void solve() {
  int n;
  string a, b;
  cin >> n >> a >> b;
  for (char &ch : a) ch -= '0';
  for (char &ch : b) ch -= '0';
  vector<int> out;
  bool flip = 0;
  int l = 0, r = n - 1;
  int idx = n;
  while (l < r) {
    idx--;
    if (flip == 0) {
      if ((a[r] ^ flip) == b[idx]) {
        r--;
        continue;
      }
      if ((a[l] ^ flip) == b[idx]) {
        out.push_back(1);
        a[l] ^= 1;
      }
      out.push_back(idx + 1);
      l++;
      flip ^= 1;
    } else {
      if ((a[l] ^ flip) == b[idx]) {
        l++;
        continue;
      }
      if ((a[r] ^ flip) == b[idx]) {
        out.push_back(1);
        a[r] ^= 1;
      }
      out.push_back(idx + 1);
      r--;
      flip ^= 1;
    }
  }
  if ((a[l] ^ flip) != b[0]) out.push_back(1);
  cout << (int)(out.size()) << ' ';
  for (auto &it : out) cout << it << ' ';
  cout << "\n";
}
int main() {
  run();
  int T;
  cin >> T;
  while (T--) {
    solve();
  }
}