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 long long MOD = 1000000007;
int dx[] = {1, 1, 0, -1, -1, -1, 0, 1};
int dy[] = {0, 1, 1, 1, 0, -1, -1, -1};
long long POW(long long x, long long y) {
  long long ans = 1;
  long long base = x;
  while (y) {
    if (y & 1) {
      ans *= base;
    }
    base = base * base;
    y = y >> 1;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long n, i, j, k, t, temp, tc;
  cin >> tc;
  while (tc--) {
    cin >> n;
    string a, b;
    vector<long long> ans;
    cin >> a >> b;
    for (i = n - 1; i >= 0; i--) {
      if (a[i] != b[i]) {
        if (a[0] != b[i]) {
          ans.push_back(i);
        } else {
          ans.push_back(0);
          if (b[i] == '1') {
            a[0] = '0';
          } else {
            a[0] = '1';
          }
          ans.push_back(i);
        }
        for (j = 0; j <= i; j++) {
          if (a[j] == '1') {
            a[j] = '0';
          } else {
            a[j] = '1';
          }
        }
        long long lb = 0, ub = i;
        while (lb < ub) {
          auto ch = a[lb];
          a[lb] = a[ub];
          a[ub] = ch;
          lb++;
          ub--;
        }
      }
    }
    cout << ans.size() << " ";
    for (auto x : ans) {
      cout << x + 1 << " ";
    }
    cout << endl;
  }
  return 0;
}