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 <class T>
void printvec(vector<T>& x, int a, int b) {
  for (int i = (a); i <= (b); i++) cout << x[i] << " ";
  cout << "\n";
}
mt19937 mt_rng(chrono::steady_clock::now().time_since_epoch().count());
const double PI = acos((double)-1);
const double EPSILON = 1e-7;
const int N = 1e6 + 1;
const int NN = 1e3 + 1;
const long long MOD = 1e9 + 7;
const long long oo = LLONG_MAX;
const int BASE = 10000;
const int di[4] = {-1, 0, 1, 0};
const int dj[4] = {0, 1, 0, -1};
const char dr[4] = {'U', 'R', 'D', 'L'};
void query() {
  int n;
  cin >> n;
  string a, b;
  cin >> a >> b;
  vector<int> res;
  for (int i = 0; i < n - 1; i++) {
    if (a[i] != a[i + 1]) {
      res.push_back(i);
    }
  }
  string s;
  for (int i = 0; i < n; i++) {
    s += a[n - 1];
  }
  int cur = a[n - 1];
  for (int i = n - 1; i >= 0; i--) {
    if (cur != b[i]) {
      res.push_back(i);
      cur = char(1 - (cur - '0') + '0');
    }
  }
  cout << (int)(res).size() << " ";
  for (int x : res) {
    cout << x + 1 << " ";
  }
  cout << "\n";
}
void solve() {
  int t;
  cin >> t;
  while (t--) {
    query();
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cout << fixed << setprecision(4);
  solve();
  cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
  return 0;
}