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;
template <typename T>
using min_queue = priority_queue<T, vector<T>, greater<T>>;
const long long MOD = 1e9 + 7;
int N;
int a[100100];
int b[100100];
vector<int> res;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int T;
  cin >> T;
  while (T-- > 0) {
    cin >> N;
    for (int i = 0; i < N; i++) {
      char c;
      cin >> c;
      a[i] = c - '0';
    }
    for (int i = 0; i < N; i++) {
      char c;
      cin >> c;
      b[i] = c - '0';
    }
    int i = 0, j = N;
    int fl = 0;
    res.clear();
    while (i < j) {
      int x, y;
      if (fl == 0) {
        x = a[i];
      } else {
        x = !a[j - 1];
      }
      y = b[j - i - 1];
      if (x == y) {
        res.push_back(1);
      }
      res.push_back(j - i);
      fl = !fl;
      if (fl == 0) {
        j--;
      } else {
        i++;
      }
    }
    cout << res.size() << ' ';
    for (int k : res) cout << k << ' ';
    cout << '\n';
  }
  return 0;
}