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;
long long INF = 1 << 28;
const double PI = acos(-1.0);
int dir[4][2] = {1, 0, -1, 0, 0, -1, 0, 1};
int knight[8][2] = {1, 2, 1, -2, 2, 1, 2, -1, -1, 2, -1, -2, -2, 1, -2, -1};
const long double EPS = 1e-7;
struct compare {
  bool operator()(const pair<int, pair<int, int> >& l,
                  const pair<int, pair<int, int> >& r) {
    if (l.first == r.first) {
      return l.second.first > r.second.first;
    }
    return l.first < r.first;
  }
};
const int mx = 500150;
int mod = 1000000007;
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n;
    scanf("%d", &n);
    string a, b;
    cin >> a >> b;
    vector<int> ans;
    int i = n - 1;
    int c = 0;
    int ekhn = n;
    while (i >= 0) {
      c = 0;
      if (a[i] == b[i]) {
        ekhn--;
        i--;
        continue;
      }
      if (a[i] == '0' && b[i] == '1') {
        c++;
      } else if (a[i] == '1' && b[i] == '0') {
        c++;
      }
      i--;
      ans.push_back(ekhn);
      ans.push_back(c);
      ans.push_back(ekhn);
      ekhn -= c;
    }
    cout << ans.size();
    for (int i = 0; i < ans.size(); i++) printf(" %d", ans[i]);
    cout << endl;
  }
  return 0;
}